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Need Help Solving Those Dreaded Word Problems Involving Quadratic Equations?

Yes, I know it's tough. You've finally mastered factoring and using the quadratic formula and now you are asked to solve more problems!

Except these are even more tough. Now you have to figure out what the problem even means before trying to solve it. I completely understand and here's where I am going to try to help!

There are many types of problems that can easily be solved using your knowledge of quadratic equations. You may come across problems that deal with money and predicted incomes (financial) or problems that deal with physics such as projectiles. You may also come across construction type problems that deal with area or geometry problems that deal with right triangles.

Lucky for you, you can solve the quadratic equations, now you just have to learn how to apply this useful skill.

On this particular page, we are going to take a look at a physics "projectile problem".

Projectiles - Example 1

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. Find the maximum height attained by the ball.

Let's first take a minute to understand this problem and what it means. We know that a ball is being shot from a cannon. So, in your mind, imagine a cannon firing a ball. We know that the ball is going to shoot from the cannon, go into the air, and then fall to the ground.

So, here's a mathematical picture that I see in my head.

Now let's talk about what each part of this problem means. In our equation that we are given we must be given the value for the force of gravity (coefficient of t 2 ). We must also use our upward velocity (coefficient of t) and our original height of the cannon/ball (the constant or 1.5). Take a look...

Now that you have a mental picture of what's happening and you understand the formula given, we can go ahead and solve the problem.

  • First, ask yourself, "What am I solving for?" "What do I need to find?" You are asked to find the maximum height (go back and take a look at the diagram). What part of the parabola is this? Yes, it's the vertex! We will need to use the vertex formula and I will need to know the y coordinate of the vertex because it's asking for the height.
  • Next Step: Solve! Now that I know that I need to use the vertex formula, I can get to work.

Just as simple as that, this problem is solved.

Let's not stop here. Let's take this same problem and put a twist on it. There are many other things that we could find out about this ball!

Projectiles - Example 2

Same problem - different question. Take a look...

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long did it take for the ball to reach the ground?

Now, we've changed the question and we want to know how long did it take the ball to reach the ground.

What ground, you may ask. The problem didn't mention anything about a ground. Let's take a look at the picture "in our mind" again.

Do you see where the ball must fall to the ground. The x-axis is our "ground" in this problem. What do we know about points on the x-axis when we are dealing with quadratic equations and parabolas?

Yes, the points on the x-axis are our "zeros" or x-intercepts. This means that we must solve the quadratic equation in order to find the x-intercept.

Let's do it! Let's solve this equation. I'm thinking that this may not be a factorable equation. Do you agree? So, what's our solution?

Hopefully, you agree that we can use the quadratic formula to solve this equation.

The first time doesn't make sense because it's negative. This is the calculation for when the ball was on the ground initially before it was shot.

This actually never really occurred because the ball was shot from the cannon and was never shot from the ground. Therefore, we will disregard this answer.

The other answer was 2.54 seconds which is when the ball reached the ground (x-axis) after it was shot. Therefore, this is the only correct answer to this problem.

Ok, one more spin on this problem. What would you do in this case?

Projectiles - Example 3

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long does it take the ball to reach a height of 20 feet?

Yes, this problem is a little trickier because the question is not asking for the maximum height (vertex) or the time it takes to reach the ground (zeros), instead it it asking for the time it takes to reach a height of 20 feet.

Since the ball reaches a maximum height of 26.5 ft, we know that it will reach a height of 20 feet on the way up and on the way down.

Let's just estimate on our graph and also make sure that we get this visual in our head.

From looking at this graph, I would estimate the times to be about 0.7 sec and 1.9 sec. Do you see how the ball will reach 20 feet on the way up and on the way down?

Now, let's find the actual values. Where will we substitute 20 feet?

Yes, we must substitute 20 feet for h(t) because this is the given height. We will now be solving for t using the quadratic formula. Take a look.

Our actual times were pretty close to our estimates. Just don't forget that when you solve a quadratic equation, you must have the equation set equal to 0. Therefore, we had to subtract 20 from both sides in order to have the equation set to 0.

You've now seen it all when it comes to projectiles!

Great Job! Hopefully you've been able to understand how to solve problems involving quadratic equations. I also hope that you better understand these common velocity equations and how to think about what this problem looks like graphically in order to help you to understand which process or formula to use in order to solve the problem.

  • Quadratic Equations
  • Projectile Problems

math word problem involving quadratic equation

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Chapter 10: Quadratics

10.7 Quadratic Word Problems: Age and Numbers

Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you will end up constructing a quadratic equation. To find the solution, you will be required to either factor the quadratic equation or use substitution.

Example 10.7.1

The sum of two numbers is 18, and the product of these two numbers is 56. What are the numbers?

First, we know two things:

[latex]\begin{array}{l} \text{smaller }(S)+\text{larger }(L)=18\Rightarrow L=18-S \\ \\ S\times L=56 \end{array}[/latex]

Substituting [latex]18-S[/latex] for [latex]L[/latex] in the second equation gives:

[latex]S(18-S)=56[/latex]

Multiplying this out gives:

[latex]18S-S^2=56[/latex]

Which rearranges to:

[latex]S^2-18S+56=0[/latex]

Second, factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} S^2&-&18S&+&56&=&0 \\ (S&-&4)(S&-&14)&=&0 \\ \\ &&&&S&=&4, 14 \end{array}[/latex]

[latex]\begin{array}{l} S=4, L=18-4=14 \\ \\ S=14, L=18-14=4 \text{ (this solution is rejected)} \end{array}[/latex]

Example 10.7.2

The difference of the squares of two consecutive even integers is 68. What are these numbers?

The variables used for two consecutive integers (either odd or even) is [latex]x[/latex] and [latex]x + 2[/latex]. The equation to use for this problem is [latex](x + 2)^2 - (x)^2 = 68[/latex]. Simplifying this yields:

[latex]\begin{array}{rrrrrrrrr} &&(x&+&2)^2&-&(x)^2&=&68 \\ x^2&+&4x&+&4&-&x^2&=&68 \\ &&&&4x&+&4&=&68 \\ &&&&&-&4&&-4 \\ \hline &&&&&&\dfrac{4x}{4}&=&\dfrac{64}{4} \\ \\ &&&&&&x&=&16 \end{array}[/latex]

This means that the two integers are 16 and 18.

Example 10.7.3

The product of the ages of Sally and Joey now is 175 more than the product of their ages 5 years prior. If Sally is 20 years older than Joey, what are their current ages?

The equations are:

[latex]\begin{array}{rrl} (S)(J)&=&175+(S-5)(J-5) \\ S&=&J+20 \end{array}[/latex]

Substituting for S gives us:

[latex]\begin{array}{rrrrrrrrcrr} (J&+&20)(J)&=&175&+&(J&+&20-5)(J&-&5) \\ J^2&+&20J&=&175&+&(J&+&15)(J&-&5) \\ J^2&+&20J&=&175&+&J^2&+&10J&-&75 \\ -J^2&-&10J&&&-&J^2&-&10J&& \\ \hline &&\dfrac{10J}{10}&=&\dfrac{100}{10} &&&&&& \\ \\ &&J&=&10 &&&&&& \end{array}[/latex]

This means that Joey is 10 years old and Sally is 30 years old.

For Questions 1 to 12, write and solve the equation describing the relationship.

  • The sum of two numbers is 22, and the product of these two numbers is 120. What are the numbers?
  • The difference of two numbers is 4, and the product of these two numbers is 140. What are the numbers?
  • The difference of two numbers is 8, and the sum of the squares of these two numbers are 320. What are the numbers?
  • The sum of the squares of two consecutive even integers is 244. What are these numbers?
  • The difference of the squares of two consecutive even integers is 60. What are these numbers?
  • The sum of the squares of two consecutive even integers is 452. What are these numbers?
  • Find three consecutive even integers such that the product of the first two is 38 more than the third integer.
  • Find three consecutive odd integers such that the product of the first two is 52 more than the third integer.
  • The product of the ages of Alan and Terry is 80 more than the product of their ages 4 years prior. If Alan is 4 years older than Terry, what are their current ages?
  • The product of the ages of Cally and Katy is 130 less than the product of their ages in 5 years. If Cally is 3 years older than Katy, what are their current ages?
  • The product of the ages of James and Susan in 5 years is 230 more than the product of their ages today. What are their ages if James is one year older than Susan?
  • The product of the ages (in days) of two newborn babies Simran and Jessie in two days will be 48 more than the product of their ages today. How old are the babies if Jessie is 2 days older than Simran?

Example 10.7.4

Doug went to a conference in a city 120 km away. On the way back, due to road construction, he had to drive 10 km/h slower, which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

The first equation is [latex]r(t) = 120[/latex], which means that [latex]r = \dfrac{120}{t}[/latex] or [latex]t = \dfrac{120}{r}[/latex].

For the second equation, [latex]r[/latex] is 10 km/h slower and [latex]t[/latex] is 2 hours longer. This means the second equation is [latex](r - 10)(t + 2) = 120[/latex].

We will eliminate the variable [latex]t[/latex] in the second equation by substitution:

[latex](r-10)(\dfrac{120}{r}+2)=120[/latex]

Multiply both sides by [latex]r[/latex] to eliminate the fraction, which leaves us with:

[latex](r-10)(120+2r)=120r[/latex]

Multiplying everything out gives us:

[latex]\begin{array}{rrrrrrrrr} 120r&+&2r^2&-&1200&-&20r&=&120r \\ &&2r^2&+&100r&-&1200&=&120r \\ &&&-&120r&&&&-120r \\ \hline &&2r^2&-&20r&-&1200&=&0 \end{array}[/latex]

This equation can be reduced by a common factor of 2, which leaves us with:

[latex]\begin{array}{rrl} r^2-10r-600&=&0 \\ (r-30)(r+20)&=&0 \\ r&=&30\text{ km/h or }-20\text{ km/h (reject)} \end{array}[/latex]

Example 10.7.5

Mark rows downstream for 30 km, then turns around and returns to his original location. The total trip took 8 hr. If the current flows at 2 km/h, how fast would Mark row in still water?

If we let [latex]t =[/latex] the time to row downstream, then the time to return is [latex]8\text{ h}- t[/latex].

The first equation is [latex](r + 2)t = 30[/latex]. The stream speeds up the boat, which means [latex]t = \dfrac{30}{(r + 2)}[/latex], and the second equation is [latex](r - 2)(8 - t) = 30[/latex] when the stream slows down the boat.

We will eliminate the variable [latex]t[/latex] in the second equation by substituting [latex]t=\dfrac{30}{(r+2)}[/latex]:

[latex](r-2)\left(8-\dfrac{30}{(r+2)}\right)=30[/latex]

Multiply both sides by [latex](r + 2)[/latex] to eliminate the fraction, which leaves us with:

[latex](r-2)(8(r+2)-30)=30(r+2)[/latex]

[latex]\begin{array}{rrrrrrrrrrr} (r&-&2)(8r&+&16&-&30)&=&30r&+&60 \\ &&(r&-&2)(8r&+&(-14))&=&30r&+&60 \\ 8r^2&-&14r&-&16r&+&28&=&30r&+&60 \\ &&8r^2&-&30r&+&28&=&30r&+&60 \\ &&&-&30r&-&60&&-30r&-&60 \\ \hline &&8r^2&-&60r&-&32&=&0&& \end{array}[/latex]

This equation can be reduced by a common factor of 4, which will leave us:

[latex]\begin{array}{rll} 2r^2-15r-8&=&0 \\ (2r+1)(r-8)&=&0 \\ r&=&-\dfrac{1}{2}\text{ km/h (reject) or }r=8\text{ km/h} \end{array}[/latex]

For Questions 13 to 20, write and solve the equation describing the relationship.

  • A train travelled 240 km at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hr. What was the rate of each engine?
  • Mr. Jones visits his grandmother, who lives 100 km away, on a regular basis. Recently, a new freeway has opened up, and although the freeway route is 120 km, he can drive 20 km/h faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones’s rate on both the old route and on the freeway?
  • If a cyclist had travelled 5 km/h faster, she would have needed 1.5 hr less time to travel 150 km. Find the speed of the cyclist.
  • By going 15 km per hr faster, a transit bus would have required 1 hr less to travel 180 km. What was the average speed of this bus?
  • A cyclist rides to a cabin 72 km away up the valley and then returns in 9 hr. His speed returning is 12 km/h faster than his speed in going. Find his speed both going and returning.
  • A cyclist made a trip of 120 km and then returned in 7 hr. Returning, the rate increased 10 km/h. Find the speed of this cyclist travelling each way.
  • The distance between two bus stations is 240 km. If the speed of a bus increases by 36 km/h, the trip would take 1.5 hour less. What is the usual speed of the bus?
  • A pilot flew at a constant speed for 600 km. Returning the next day, the pilot flew against a headwind of 50 km/h to return to his starting point. If the plane was in the air for a total of 7 hours, what was the average speed of this plane?

Example 10.7.6

Find the length and width of a rectangle whose length is 5 cm longer than its width and whose area is 50 cm 2 .

First, the area of this rectangle is given by [latex]L\times W[/latex], meaning that, for this rectangle, [latex]L\times W=50[/latex], or [latex](W+5)W=50[/latex].

math word problem involving quadratic equation

Multiplying this out gives us:

[latex]W^2+5W=50[/latex]

[latex]W^2+5W-50=0[/latex]

Second, we factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} W^2&+&5W&-&50&=&0 \\ (W&-&5)(W&+&10)&=&0 \\ &&&&W&=&5, -10 \\ \end{array}[/latex]

We reject the solution [latex]W = -10[/latex].

This means that [latex]L = W + 5 = 5+5= 10[/latex].

Example 10.7.7

If the length of each side of a square is increased by 6, the area is multiplied by 16. Find the length of one side of the original square.

math word problem involving quadratic equation

The relationship between these two is:

[latex]\begin{array}{rrl} \text{larger area}&=&16\text{ times the smaller area} \\ (x+12)^2&=&16(x)^2 \end{array}[/latex]

Simplifying this yields:

[latex]\begin{array}{rrrrrrr} x^2&+&24x&+&144&=&16x^2 \\ -16x^2&&&&&&-16x^2 \\ \hline -15x^2&+&24x&+&144&=&0 \end{array}[/latex]

Since this is a problem that requires factoring, it is easiest to use the quadratic equation:

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=-15, b=24\text{ and }c=144[/latex]

Substituting these values in yields [latex]x = 4[/latex] or [latex]x=-2.4[/latex] (reject).

Example 10.7.8

Nick and Chloe want to surround their 60 by 80 cm wedding photo with matting of equal width. The resulting photo and matting is to be covered by a 1 m 2 sheet of expensive archival glass. Find the width of the matting.

math word problem involving quadratic equation

[latex](L+2x)(W+2x)=1\text{ m}^2[/latex]

[latex](80\text{ cm }+2x)(60\text{ cm }+2x)=10,000\text{ cm}^2[/latex]

[latex]4800+280x+4x^2=10,000[/latex]

[latex]4x^2+280x-5200=0[/latex]

Which reduces to:

[latex]x^2 + 70x - 1300 = 0[/latex]

Second, we factor this quadratic to get our solution.

It is easiest to use the quadratic equation to find our solutions.

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=1, b=70\text{ and }c=-1300[/latex]

Substituting the values in yields:

[latex]x=\dfrac{-70\pm \sqrt{70^2-4(1)(-1300)}}{2(1)}\hspace{0.5in}x=\dfrac{-70\pm 10\sqrt{101}}{2}[/latex]

[latex]x=-35+5\sqrt{101}\hspace{0.75in} x=-35-5\sqrt{101}\text{ (rejected)}[/latex]

For Questions 21 to 28, write and solve the equation describing the relationship.

  • Find the length and width of a rectangle whose length is 4 cm longer than its width and whose area is 60 cm 2 .
  • Find the length and width of a rectangle whose width is 10 cm shorter than its length and whose area is 200 cm 2 .
  • A large rectangular garden in a park is 120 m wide and 150 m long. A contractor is called in to add a brick walkway to surround this garden. If the area of the walkway is 2800 m 2 , how wide is the walkway?
  • A park swimming pool is 10 m wide and 25 m long. A pool cover is purchased to cover the pool, overlapping all 4 sides by the same width. If the covered area outside the pool is 74 m 2 , how wide is the overlap area?
  • In a landscape plan, a rectangular flowerbed is designed to be 4 m longer than it is wide. If 60 m 2 are needed for the plants in the bed, what should the dimensions of the rectangular bed be?
  • If the side of a square is increased by 5 units, the area is increased by 4 square units. Find the length of the sides of the original square.
  • A rectangular lot is 20 m longer than it is wide and its area is 2400 m 2 . Find the dimensions of the lot.
  • The length of a room is 8 m greater than its width. If both the length and the width are increased by 2 m, the area increases by 60 m 2 . Find the dimensions of the room.

Answer Key 10.7

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Last modified on August 3rd, 2023

Quadratic Equation Word Problems

Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them:

  • By Completing the Square
  • By Factoring
  • By Quadratic Formula
  • By graphing

For each process, follow the following typical steps:

  • Make the equation
  • Solve for the unknown variable using the appropriate method
  • Interpret the result

The product of two consecutive integers is 462. Find the numbers?

Let the numbers be x and x + 1 According to the problem, x(x + 1) = 483 => x 2 + x – 483 = 0 => x 2 + 22x – 21x – 483 = 0 => x(x + 22) – 21(x + 22) = 0 => (x + 22)(x – 21) = 0 => x + 22 = 0 or x – 21 = 0 => x = {-22, 21} Thus, the two consecutive numbers are 21 and 22.

The product of two consecutive positive odd integers is 1 less than four times their sum. What are the two positive integers.

Let the numbers be n and n + 2 According to the problem, => n(n + 2) = 4[n + (n + 2)] – 1 => n 2 + 2n = 4[2n + 2] – 1 => n 2 + 2n = 8n + 7 => n 2 – 6n – 7 = 0 => n 2 -7n + n – 7 = 0 => n(n – 7) + 1(n – 7) = 0 => (n – 7) (n – 1) = 0 => n – 7 = 0 or n – 1 = 0 => n = {7, 1} If n = 7, then n + 2 = 9 If n = 1, then n + 1 = 2 Since 1 and 2 are not possible. The two numbers are 7 and 9

A projectile is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet tower. If height after t seconds is reprented by h(t) = -16t 2 + 64t + 96. Find the maximum height the projectile reaches. Also, find the time it takes to reach the highest point.

Since the graph of the given function is a parabola, it opens downward because the leading coefficient is negative. Thus, to get the maximum height, we have to find the vertex of this parabola. Given the function is in the standard form h(t) = a 2 x + bx + c, the formula to calculate the vertex is: Vertex (h, k) = ${\left\{ \left( \dfrac{-b}{2a}\right) ,h\left( -\dfrac{b}{2a}\right) \right\}}$ => ${\dfrac{-b}{2a}=\dfrac{-64}{2\times \left( -16\right) }}$ = 2 seconds Thus, the time the projectile takes to reach the highest point is 2 seconds ${h\left( \dfrac{-b}{2a}\right)}$ = h(2) = -16(2) 2 – 64(2) + 80 = 144 feet Thus, the maximum height the projectile reaches is 144 feet

The difference between the squares of two consecutive even integers is 68. Find the numbers.

Let the numbers be x and x + 2 According to the problem, (x + 2) 2 – x 2 = 68 => x 2 + 4x + 4 – x 2 = 68 => 4x + 4 = 68 => 4x = 68 – 4 => 4x = 64 => x = 16 Thus the two numbers are 16 and 18

The length of a rectangle is 5 units more than twice the number. The width is 4 unit less than the same number. Given the area of the rectangle is 15 sq. units, find the length and breadth of the rectangle.

Let the number be x Thus, Length = 2x + 5 Breadth = x – 4 According to the problem, (2x + 5)(x – 4) = 15 => 2x 2 – 8x + 5x – 20 – 15 = 0 => 2x 2 – 3x – 35 = 0 => 2x 2 – 10x + 7x – 35 = 0 => 2x(x – 5) + 7(x – 5) = 0 => (x – 5)(2x + 7) = 0 => x – 5 = 0 or 2x + 7 = 0 => x = {5, -7/2} Since we cannot have a negative measurement in mensuration, the number is 5 inches. Now, Length = 2x + 5 = 2(5) + 5 = 15 inches Breadth = x – 4 = 15 – 4 = 11 inches

A rectangular garden is 50 cm long and 34 cm wide, surrounded by a uniform boundary. Find the width of the boundary if the total area is 540 cm².

Given, Length of the garden = 50 cm Width of the garden = 34 cm Let the uniform width of the boundary be = x cm According to the problem, (50 + 2x)(34 + 2x) – 50 × 34 = 540 => 4x 2 + 168x – 540 = 0 => x 2 + 42x – 135 = 0 Since, this quadratic equation is in the standard form ax 2 + bx + c, we will use the quadratic formula, here a = 1, b = 42, c = -135 x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$ => ${\dfrac{-42\pm \sqrt{\left( 42\right) ^{2}-4\times 1\times \left( -135\right) }}{2\times 1}}$ => ${\dfrac{-42\pm \sqrt{1764+540}}{2}}$ => ${\dfrac{-42\pm \sqrt{2304}}{2}}$ => ${\dfrac{-42\pm 48}{2}}$ => ${\dfrac{-42+48}{2}}$ and ${\dfrac{-42-48}{2}}$ => x = {-45, 3} Since we cannot have a negative measurement in mensuration the width of the boundary is 3 cm

The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

Let the length of the other two sides be x and x + 4 According to the problem, (x + 4) 2 + x 2 = 20 2 => x 2 + 8x + 16 + x 2 = 400 => 2x 2 + 8x + 16 = 400 => 2x 2 + 8x – 384 = 0 => x 2 + 4x – 192 = 0 => x 2 + 16x – 12x – 192 = 0 => x(x + 16) – 12(x + 16) = 0 => (x + 16)(x – 12) = 0 => x + 16 = 0 and x – 12 = 0 => x = {-16, 12} Since we cannot have a negative measurement in mensuration, the lengths of the sides are 12 and 16

Jennifer jumped off a cliff into the swimming pool. The function h can express her height as a function of time (t) = -16t 2 +16t + 480, where t is the time in seconds and h is the height in feet. a) How long did it take for Jennifer to attain a maximum length. b) What was the highest point that Jennifer reached. c) Calculate the time when Jennifer hit the water?

Comparing the given function with the given function f(x) = ax 2 + bx + c, here a = -16, b = 16, c = 480 a) Finding the vertex will give us the time taken by Jennifer to reach her maximum height  x = ${-\dfrac{b}{2a}}$ = ${\dfrac{-16}{2\left( -16\right) }}$ = 0.5 seconds Thus Jennifer took 0.5 seconds to reach her maximum height b) Putting the value of the vertex by substitution in the function, we get ${h\left( \dfrac{1}{2}\right) =-16\left( \dfrac{1}{2}\right) ^{2}+16\left( \dfrac{1}{2}\right) +480}$ => ${-16\left( \dfrac{1}{4}\right) +8+480}$ => 484 feet Thus the highest point that Jennifer reached was 484 feet c) When Jennifer hit the water, her height was 0 Thus, by substituting the value of the height in the function, we get -16t 2 +16t + 480 = 0 => -16(t 2 + t – 30) = 0 => t 2 + t – 30 = 0 => t 2 + 6t – 5t – 30 = 0 => t(t + 6) – 5(t + 6) = 0 => (t + 6)(t – 5) = 0 => t + 6 = 0 or t – 5 = 0 => x = {-6, 5} Since time cannot have any negative value, the time taken by Jennifer to hit the water is 5 seconds.

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Quadradic Equation Word Problems - Examples & Practice - Expii

Quadradic equation word problems - examples & practice, explanations (3), quadratic word problems.

Let's look at 2 pretty common types of word problems that use quadratic functions. Recall that quadratic expressions follow this general form: y=ax2+bx+c In a quadratic expression, a and b are coefficients (numbers in front of the variable x), and c is constant (a number by itself). It's important to remember that a≠0 . It's possible for b=0 and/or c=0, but they don't have to.

Example 1: Word problems involving area

Many quadratic word problems use the area of a rectangle. Here is a graphic to get you started

math word problem involving quadratic equation

Image source: By Caroline Kulczycky

Here is another example.

Singh knows the total area of his garden is 50 m2, but he doesn't know the length or width. Use the diagram below to find the length and width in meters (m).

A rectangle with width (x - 5) and length x.

Recall that the formula for area of a rectangle is length⋅width. So, let's plug the values from Singh's garden into that formula, and set it equal to 50 (the area). area=length⋅width50=x⋅(x−5)

Now, let's multiply out this equation and get it in the standard quadratic form (written at the top of this explanation):

Once we multiply the equation out and rearrange it, what does it look like (in quadratic form)?

−x2−5x−50=0

−x2−5x+50=0

Related Lessons

(videos) solving geometry word problems using quadratic equations.

by PatrickJMT

math word problem involving quadratic equation

This video by Patrick JMT covers a quadratic word problem based on area of a rectangle.

Patrick walks us through how to solve the following quadratic geometry problem:

“A picture inside a frame is 2 inches longer than it is wide. The picture is in a frame that has width 3 inches on each side of the picture. If the area of the pic, including the frame is 195 in.2, find the dimensions of the frame.”

He goes on to explain how we get the equations for the sides of the frame, so the whole problem looks like:

A picture within a picture frame.  The picture has dimensions of x and x+2, and the picture frame has dimensions of x+6 and x+8

We know the area of the picture frame, 195 in2. We can set the two sides of the frame equal to that and find a value for x.

(x+6)(x+8)=195

Using the FOIL method , we get an equation that looks like

x2+14x+48=195

We want the left side to equal 0, so subtracting 195 from both sides will give us

x2+14x–147=0

This looks pretty hard to factor. Let’s use our handy dandy quadratic formula .

Our values are going to be a=1,b=14, and c=−147

x=−b±√b2−4ac2ax=−14±√142–4(1)(−147)2(1)x=−14±√196+5882x=−14±√7842x=−14±282x=−7±14

*Note: be careful with the square root!

This leaves us two values for x, x=7 or x=−21. We cannot have a negative measurement, so our final answer is x=7.

(Videos) Solve Projectile Problems Using Quadratics

math word problem involving quadratic equation

This video by Patrick JMT works through a word problem based on projectile motion .

Solving a Projectile Problem Using Quadratics

Patrick uses the following problem as an example of projectile motion:

“Larry throws a rock in the air. The height, h, in feet above the ground of the rock is given by: h=−16t2+123t+40. How long is the rock in the air?”

From this equation, at time t=0, we gather that h=40. This means Larry is on a cliff with height 40. We want to know the time at which the rock reaches a height of 0. So, plugging in 0 for h, we get an equation:

0=−16t2+123t+40

This is now a quadratic equation, which we can solve for using the quadratic formula! We have values of a=16,b=123, and c=40. We can plug these into the quadratic formula: t=−b±√b2−4ac2at=−123±√(123)2−4(−16)(40)2(−16)t=−123±√17689−32t=123±√1768932t=123±13332t=8,−516

We get two solutions: 8 and −516. t is time, which can't be negative, so our answer is t=8.

Quadratic Equations Word Problems

These lessons, with videos, examples, and step-by-step solutions, help Algebra 1 students learn to solve geometry word problems using quadratic equations.

Related Pages Solving Quadratic Equations by Factoring Solving Quadratic Equations by Completing the Square More Lessons for Grade 9 Math Worksheets

Quadratic equations - Solving word problems using factoring of trinomials Question 1a: Find two consecutive integers that have a product of 42

Quadratic equations - Solving word problems using factoring of trinomials Question 1b: There are three consecutive integers. The product of the two larger integers is 30. Find the three integers.

Quadratic Equations - Solving Word problems by Factoring Question 1c: A rectangular building is to be placed on a lot that measures 30 m by 40 m. The building must be placed in the lot so that the width of the lawn is the same on all four sides of the building. Local restrictions state that the building cannot occupy any more than 50% of the property. What are the dimensions of the largest building that can be built on the property?

More Word Problems Using Quadratic Equations Example 1 Suppose the area of a rectangle is 114.4 m 2 and the length is 14 m longer than the width. Find the length and width of the rectangle.

More Word Problems Using Quadratic Equations Example 2 A manufacturer develops a formula to determine the demand for its product depending on the price in dollars. The formula is D = 2,000 + 100P - 6P 2 where P is the price per unit, and D is the number of units in demand. At what price will the demand drop to 1000 units?

More Word Problems Using Quadratic Equations Example 3 The length of a car’s skid mark in feet as a function of the car’s speed in miles per hour is given by l(s) = .046s 2 - .199s + 0.264 If the length of skid mark is 220 ft, find the speed in miles per hour the car was traveling.

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SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

Problem 1 :

If the difference between a number and its reciprocal is ²⁴⁄₅ , find the number.

Let x be the required number. Then its reciprocal is  ¹⁄ₓ be its reciprocal.

5 (x 2  - 1) = 24x

5 x 2  - 5 = 24x

5x 2  - 24x - 5 = 0

5x 2  - 25x + 1x - 5 = 0

5x(x - 5) + 1(x - 5) = 0

(5x + 1)(x - 5) = 0

5x + 1 = 0  or  x - 5 = 0

x = - ⅕   or  x = 5

So, the required number is - ⅕  or -5.

Problem 2 :

A garden measuring 12m by 16m is to have a pedestrian pathway that is w meters wide installed all the way around so that it increases the total area to 285 m 2 . What is the width of the pathway?

10thnewsylabusex3.12q2

From the picture above, length of the garden including pathway is (12 + 2w) and width is (16 + 2w).

Total Area = 285 m 2

Length  ⋅ Width = 285

(12 + 2w) (16 + 2w) = 285

  192 + 24w + 32w + 4w 2  = 285

 4w 2 + 56w + 192 - 285 = 0

4w 2  + 56w - 93 = 0

Solve the above quadratic equation using quadratic formula.

w = -15.5  or  w = 1.5 

Since w is the width of the pathway, it can not be negative. So, w = 1.5.

Therefore , the width of the pathway is 1.5 m .

Problem 3 :

A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the bus.

Distance covered = 90 km

Let x be the original speed  of the bus.

Increased speed = x + 15

Time taken by the bus in original speed :

Time taken by the bus in increased speed :

= ⁹⁰⁄₍ₓ ₊ ₁₅₎

From the given information, the difference between and  ⁹⁰⁄ₓ  and ⁹⁰⁄₍ₓ ₊ ₁₅₎  is 30 minutes or ½ hour .

2700 = x 2  + 15x 

x 2  + 15x - 2700 = 0

x 2  + 60x - 45x - 2700 = 0

x(x + 60) - 45(x + 60) = 0

(x - 45)(x + 60) = 0

x = 45  or  x = -60

Since x represents the original speed of the bus, it can never be negative. So, x = 45.

Therefore, the original speed of the bus is 45 km per hour.

Problem 4 :

John and Jivanti together had 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. Find the number marbles each one them had initially.

Let x be the number of marbles that john has.

Then, the number of marbles that Jivanthi has

After losing 5 marbles, 

number of marbles with John = x - 5

number of marbles with Jivanti = 45 - x - 5 = 40 - x

The product of number of marbles = 124

(x - 5)(40 - x) = 124

40x - x 2  - 200 + 5x = 124

45x - x 2  - 200 = 124

x 2  - 45x + 124 + 200 = 0

x 2  - 45x + 324 = 0

x 2 - 36x - 9x + 324 = 0

x(x - 36) - 9(x - 36) = 0

(x - 9)(x - 36) = 0

x - 9 = 0  or  x - 36 = 0

x = 9  or  x = 36

If x = 9, 

If x = 36, 

So, John had 36 marbles, when Jivanti had 9 marbles or Jivanti had 36 marbles, when John had 9 marbles. 

Problem 5 :

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in dollars) was found to be 55 minus the number of toys produced in a day, the total cost of production was $750. Find the number of toys produced on that day.

Let x be the number of toys produced in a particular day.

Cost of production  of one toy (in dollars) :

Total cost =  Number of toys  ⋅  Cost of one toy

750 = x (55 - x)

750 = 55x - x 2

x 2  - 55x + 750 = 0

x 2  - 30x - 25x + 750 = 0

x(x - 30) - 25(x - 30) = 0

(x - 30)(x - 25) = 0

  x - 30 = 0  or  x - 25 = 0

   x = 30  or  x = 25

So, the number of toys produced on that particular day is 30 or 25.

Problem 6 :

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Let x be the length of shorter side of the rectangle

length of diagonal = x + 60

length of longer side = x + 30

6thquestion

Using Pythagorean Theorem in the right triangle ABC,

(x + 60) 2  = x 2  + (x + 30 ) 2

x 2  + 60 2  + 2(x)(60) = x 2 + x 2  + 2(x)(30) + 30 2

x 2  + 3600 + 120x = 2x 2  + 60x + 900

2 x 2  - x 2  + 60x - 120x + 900 - 3600 = 0

x 2  - 60x - 2700 = 0

x 2  - 90x + 30x - 2700 = 0

x(x - 90) + 30(x - 90) = 0

 (x + 30)(x - 90) = 0

  x + 30 = 0  or  x - 90 = 0

   x = -30  or  x = 90

Because x represents breadth of the rectangle, it can never be negative. So,  x = 90.

Therefore, 

length of the shorter side = 90 m

length of the longer side = 90 + 30 = 120 m

Problem 7 :

The difference of squares of two positive numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let x be the larger number and  y be the smaller number.

The square of the smaller number is 8 times the larger number.

y 2  = 8x -----(1)

The difference of squares of two numbers is 180.

 x 2  - y 2  = 180

Substitute  y 2  = 8x and solve for x.

    x 2  - 8x = 180

x 2  - 8x - 180 = 0

    x 2  - 18x + 10x - 180 = 0

x(x - 18) + 10(x - 18) = 0

 (x - 18)(x + 10) = 0 

 x - 18 = 0  or  x + 10 = 0

 x = 18  or  x = -10

Because the numbers are positive, x = -10 can not be accepted. So,  x = 18.

Substitute x = 18 into (1).

y 2  = 8(18)

y 2  = 144

 y =  √144

 y = 12

Therefore, the larger number is 18  and the smaller number is 12.

Problem 8 :

A train travels 360 miles at a uniform speed. If the speed had been 5 miles/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Distance covered = 360 miles

Let x be the original speed of the train.

If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey.

Increased speed = x + 5

Time taken by the train in original speed :

=  ³⁶⁰⁄ₓ

Time taken by the train in increased speed :

=  ³⁶⁰⁄₍ₓ ₊ ₅₎

From the given information, the difference between and  ³⁶⁰⁄ₓ   and ³⁶⁰⁄₍ₓ ₊ ₅₎  is 1  hour .

1800 = x 2  + 5x

x 2  + 5x - 1800 = 0

x 2  - 40x + 45x - 1800 = 0

x(x - 40) + 45(x - 40) = 0

(x - 40)(x + 45) = 0

 x - 40 = 0  or  x + 45 = 0

 x = 40  or  x = -45

Since x represents the original speed of the train, it can never be negative. So, x = 40.

Therefore, the original speed of the train is 40 miles/hr.

Problem 9 :

Find two consecutive positive even integers whose squares have the sum 340.

Let x and ( x + 2) be the two positive even integers.

Sum of their squares = 340

x 2 + (x + 2) 2  = 340

x 2 + (x + 2)(x + 2) = 340

x 2 + x 2  + 2x + 2x + 4 = 340

2x 2 + 4x + 4 = 340

2x 2 + 4x - 336 = 0

Divide both sides by 2.

x 2 + 2x - 168 = 0

Solve by factoring.

x 2  - 12x + 14x - 168 = 0

x(x - 12) + 14(x - 12) = 0

(x - 12)(x + 14) = 0

x - 12 = 0  or  x + 14 = 0

x = 12  or  x = -14

Since the integers are positive, x can not be negative.

Therefore, the two positive even integers are 12 and 14.

Problem 10 :

The sum of squares of three consecutive natural numbers is 194. Determine the numbers.

Let x , x + 1 and x + 2 be three consecutive natural numbers.

Sum of their squares = 194

x 2  + (x + 1) 2  + (x + 2) 2  = 194

x 2  + (x + 1)(x + 1) + (x + 2)(x + 2) = 194

x 2  + x 2  + x + x + 1 + x 2  + 2x + 2x + 4 = 194

3x 2  + 6x + 5 = 194

 3x 2  + 6x - 189 = 0

Divide both sides by 3.

x 2  + 2x - 63 = 0

x 2  - 7x + 9x - 63 = 0

x(x - 7) + 9(x - 7) = 0

(x - 7)(x + 9) = 0

x - 7 = 0  or  x + 9 = 0

x = 7  or  x = -9

Since the numbers are natural numbers, x  can not be negative.

Therefore, the three  consecutive natural numbers are 7, 8 and 9.

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Word Problems with Quadratic Equations

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Basics on the topic Word Problems with Quadratic Equations

There are many ways to solve quadratic equations. You can factor them, complete the square, graph them, and use the quadratic formula, for instance.

At least one of these methods can be used to solve any problem involving a quadratic equation, and which method you choose depends on the kind of problem you are presented with.

If the quadratic equation can be represented by mapping or a table, then graphing can do the trick. If the quadratic equation is factorable, then factoring, of course, and completing the square are good choices. When the quadratic equation has coefficients that are real numbers, like decimals, fractions, or maybe even radicals, then using the quadratic formula is highly recommended.

One must note though that these problems often look for concrete answers, like units of measurement or quantity. A negative root, though a valid solution to a quadratic equation, may not be the answer that a word problem is looking for. A positive root, or possibly roots, then is the more appropriate final answer.

Analyze Functions Using Different Representations.

CCSS.MATH.CONTENT.HSF.IF.C.8.A

Transcript Word Problems with Quadratic Equations

The mighty King Wallace sits on his throne and rules his kingdom every day of the year, except for his birthday. Every year, to celebrate his birthday he takes a trip to visit various scenic outlooks and famous locations within his kingdom. For this year’s birthday celebration, the decision is difficult, there're so many destinations to choose from.But finally, the mighty king decides to visit one of the more beautiful castles in his kingdom for the umteenth time. To prepare for the king's visit, the servants must cover the ground around the castle with carpet, so the King’s shoes won’t get dirty. How much carpet do they need?

Setting up the quadratic equation

To figure this out, we can use a quadratic equation . Take a look at the diagram of the castle. The length of the area that needs carpet is equal to an unknown length, 'x', plus 9, the width is 'x' + 3, and the total area is 72 square wallacesons. You might be wondering, what’s a wallaceson? King Wallace devised his very own system of weights and measures. I know! What a narcissist but whatever, right?! Okay back to the problem. To help the servants figure out the amount of carpet they need, we can set up an equation and solve for 'x' . First use FOIL : first, outer, inner, last. You know the drill.

Factoring the quadratic equation

Next, you'll need to factor the quadratic equation. Using the standard form of the quadratic equation as your guide, set the equation equal to zero . Then find the product of ac that sums to 'b'. That's negative 3 and 15. Now set each of the two binomials factors equal to 0. You're not done yet; you still have to solve for 'x'. Use opposite operations to solve each mini equation. X has two possible solutions, 3 and negative 15.

Completing the square

But factoring’s not the only game in town. We can also solve for 'x' by completing the square . FOIL first , so it's easier to work with. To complete the square, follow these steps: Move the constant to the other side of the equal sign . Then take the 'b' value , divide it by 2, and square it, and then add this number to both sides of the equation. Factor the left side of the equal sign and finally, solve for 'x' , by taking the square root of each side of the equal sign and finish it off with PEMDAS . X is equal to 3 and -15, just like before.

Let’s plug in the values for 'x' into the equation, so the servants can order the carpet. Plug in 'x' is equal to 3. The measurements are 12 wallacesons by 6 wallacesons! That’s a big carpet! Now for the second value, 'x' = -15. The measurements of the carpet are equal to -6 and -12. You can’t have a negative measure of carpet so although 'x' = -15 works as a solution for the quadratic equation. It’s not a valid solution for this situation. So the correct solution is 'x' is equal to 3.

Indecisive as always, the king changes his mind at the very last minute and decides he wants to visit a place he’s never been to before. He'll go to a village near the border of his kingdom. He’s been curious about this village because he heard they have some unusual customs. When the King gets to the village, the villagers seem normal enough. The villagers stand with baskets of apples patiently waiting for his speech to begin. Luckily for the king, his squire knows all about the village's unusual customs, the villagers always give an enthusiastic welcome to visitors. By throwing apples at them! To keep the king safe, the squire will have to control the crowd, but how can he do that?

Graphing the parabola

The king will give his speech while standing on a tower 5 wallace yards high, at the point on the graph (0, 5). The trajectory of the apples can be described by the function of 'x' = -0.25x² + 2x + 5, since it's a quadratic equation, we know the shape of the graph is a parabola . So, where does the crowd need to stand to so they won't be able to hit the king with apples? This whole situation sounds a little crazy, right? What can you do? You have to respect peoples' customs, right?

Let’s solve this problem by graphing . Plug in a few points for 'x' and determine the corresponding 'y' values . Plot the points and then draw in the parabola . Where the graph touches the x-axis is a possible solution set for this quadratic equation, at 'x' = 10 and -2. The king will be safe as long as the crowd is placed more than 10 feet in front of the podium or 2 feet behind.

Quadratic formula

We can also calculate the solution by using the quadratic formula. Use the standard quadratic equation as your guide to determine the values for a, b, and c, but first manipulate the equation so the 'a' value is equal to positive 1. You could skip this step, but it makes the numbers easier to work with. Now substitute the 'a', 'b' and 'c' values into the quadratic formula, and then do the math. The answer is the same as before, 'x' is equal to 10 and -2. Just as the squire expected, the crowd welcomed the king by throwing lots of apples but thanks to the quadratic equation, the king gave his speech without a single apple finding its mark.

Word Problems with Quadratic Equations exercise

Explain how to solve a quadratic equation by factoring..

  • multiply the F irst $x^2$
  • multiply the O uter $9x$
  • multiply the I nner $3x$
  • multiply the L asr $27$

Here is an example for solving an equation; subtraction is the opposite operation of addition.

Use the zero factor property: if a product is equal to zero then one of its factors must also be equal to zero.

The servants have to calculate the total length of the carpet.

To do this they first use the FOIL method for multiplying the two binomials:

$(x+9)(x+3)=x^2+9x+3x+27=x^2+12x+27$.

This quadratic term is equal to the given total area. So we get

$x^2+12x+27=72$,

which is equivalent to

$\begin{array}{rcl} x^2+12x+27&=&~72\\ \color{#669900}{-72} & &\color{#669900}{-72}\\ x^2+12x-45&=&~0. \end{array}$

  • $1\times(-45)=-45$ but $1-45=-44$
  • $3\times (-15)=-45$ but $3-15=-12$
  • $-3\times 15=-45$ and $-3+15=12$ $~~~~~$✓

To get the solutions we use opposite operations twice:

$\begin{array}{rcl} x-3&=&~0\\ \color{#669900}{+3} & &\color{#669900}{+3}\\ x&=&~3 \end{array}$

$\begin{array}{rcl} x+15&=&~0\\ \color{#669900}{-15} & &\color{#669900}{-15}\\ x&=&~-15 \end{array}$

For $x=3$ we get that the length is $3+9=12$ and the width is $3+3=6$. And for $x=-15$, the length is $-15+9=-6$ ... so we can stop because there don't exist any negative lengths.

This means that both $x=3$ as well as $x=-15$ are solutions to the quadratic equation, but only $x=3$ is a possible solution in our given situation.

Determine how to solve quadratic equations graphically.

The function above is a quadratic function.

Any linear function is given by $f(x)=mx+b$, where $m$ is the slope and $b$ the $y$-intercept. The graph of such a function is a line.

The graph of a quadratic function is a parabola. Each parabola has at most two $x$-intercepts.

There exist different ways to solve quadratic equations like $ax^2+bx+c=0$.

For example, lets draw the parabola in a coordinate plane which corresponds to the graph of

  • Plug in a few values for $x$ and calculate the corresponding $y$-values.
  • Plot the resulting ordered pairs $(x,y)$ in a coordinate plane.
  • Connect those pairs to get the corresponding parabola.

Determine how to solve quadratic equations by completing the square.

  • multiply the O uter $-3x$
  • multiply the I nner $2x$
  • multiply the L asr $-6$

To complete a quadratic term $x^2+bx$ add $\left(\frac b2\right)^2$.

Here is an example for solving an equation by taking a square root.

Don't forget the $\pm$ sign in your calculation!

Check the solutions. Do they makes sense when trying to determine the positive length and width of the moat?

First we use the FOIL method for multiplying the two binomials

$(x+2)(x-3)=x^2-3x+2x-6=x^2-x-6$.

This must be equal to the given total area. So we get $x^2-x-6=50$, which is equivalent to

$\begin{array}{rcl} x^2-x-6&=&~50\\ \color{#669900}{+6} & &\color{#669900}{+6}\\ x^2-x&=&~56 \end{array}$

Now we complete the quadratic term on the left-hand side to a square. For this we add $(\frac12)^2$:

$\begin{array}{rcl} x^2-x&=&~56\\ \color{#669900}{+\left(\frac12\right)^2} & &\color{#669900}{+\left(\frac12\right)^2}\\ x^2-x+\left(\frac12\right)^2&=&~56+\left(\frac12\right)^2\\ \left(x-\frac12\right)^2&=&~56.25 \end{array}$

Next we take the square root of both sides to get

$x-\frac12=\pm7.5$.

Lastly we add $\frac12=0.5$, which leads to the desired solutions

$x=0.5+7.5=8$ or $x=0.5-7.5=-7$.

Let's check if both solutions work with the problem at hand, keeping in mind that we are trying to find the length and width of a moat.

For $x=8$ we get the length $8+2=10$ and the width $8-3=5$. Both are positive, so this solution works.

For $x=-7$ we get the length $-7+2=-5$ and the width $-7-3=-10$. A negative length and width doesn't make any sense, so this solution can't work in our given situation.

Calculate where the castle ditches have to be built by using the quadratic formula.

You can also multiply the equation $-0.5x^2-1.5x+5=0$ by $-2$ to get $x^2+3x-10=0$.

Both solutions are whole numbers. One is negative and the other one is positive.

You can solve each quadratic equation $ax^2+bx+c=0$ by using the quadratic formula,

$x=\frac{-b \pm\sqrt{b^2-4ac}}{2a}$.

First determine $a$, $b$ and $c$ in the equation and then plug those values into the quadratic formula.

For $-0.5x^2-1.5x+5=0$, we have that $a=-0.5$, $b=-1.5$, and $c=5$.

So we calculate

$\begin{array}{rcl} x&=&\frac{1.5\pm\sqrt{(-1.5)^2-4(-0.5)(5)}}{2(-0.5)}\\ &=&~\frac{1.5\pm\sqrt{2.25+10}}{-1}\\ &=&~-1.5\pm\sqrt{12.25}\\ x_1&=&~-1.5+3.5=2\\ x_2&=&~-1.5-3.5=-5 \end{array}$

The desired solutions are then $x=2$ or $x=-5$.

Perhaps you'd like to multiply the quadratic equation $-0.5x^2-1.5x+5=0$ with $-2$ to get $x^2+3x-10=0$. You don't have to do it but it makes the calculations a little bit less complicated.

Name the methods for solving quadratic equations.

The solutions of the corresponding quadratic equation are $x=10$ and $x=90$.

This is the quadratic formula.

The zero factor property states that if a product is equal to zero then one of its factors must also equal zero.

It doesn't matter at all which method you choose to find the solutions. If the solutions exist, then they are always the same.

  • You could factor the left-hand side of the equation $x^2+bx+c=0$ to $(x+d)(x+e)=0$ and get the solutions $x=-d$ and $x=-e$.
  • You can complete the square to get $(x+e)^2=d$. You can solve this equation by taking the square root of both sides.
  • You can use the quadratic formula for solving $ax^2+bx+c=0$.
  • You also could draw the parabola corresponding to $f(x)=ax^2+bx+c$. The $x$-intercepts of this parabola are the desired solutions.

Solve the following quadratic equations.

Decide which method you you like to use. No matter which method you choose, your results will always be the same.

Here you see the quadratic formula for solving quadratic equations like $ax^2+bx+c=0$.

For solving a quadratic equation $x^2+bx+c=0$ by factoring, find the factors of $c$ which sum to $b$.

Here is an example of complete the square of a quadratic equation.

Let's solve the various quadratic equations using the different methods we learned:

  • Look for the factors of $8$ which sum to $-2$, namely $2\times (-4)=-8$ and $2-4=-2$ $~~~~~$✓.
  • Thus we get the equivalent equation $(x+2)(x-4)=0$ which we solve by using opposite operations:

$\begin{array}{rcl} x+2&=&~0\\ \color{#669900}{-2} & &\color{#669900}{-2}\\ x&=&~-2 \end{array}$

$\begin{array}{rcl} x-4&=&~0\\ \color{#669900}{+4} & &\color{#669900}{+4}\\ x&=&~4 \end{array}$

This method (better!) works for whole solutions.

  • Subtract $7$ from both sides to get $x^2-8x=-7$.
  • Adding $4^2$ on both sides leads to $x^2-8x+4^2=-7+4^2$.
  • $(x-4)^2=9$ is the resulting equation.
  • Take the square root on both sides and don't forget the $\pm$ sign: $x-4=\pm3$.
  • Almost done: add $4$ to get $x=4+3=7$ or $x=4-3=1$.

You can also use the quadratic formula , which we will use to solve $2x^2+4x-6=0$. Nothing that $a=2$, $b=4$, and $c=-6$, we have that

$\begin{array}{rcl} x&=&~\frac{-4\pm\sqrt{(-4)^2-4(2)(-6)}}{2(2)}\\ &=&~\frac{-4\pm\sqrt{16+48}}{4}\\ &=&~\frac{-4\pm\sqrt{64}}{4}\\ x_1&=&~\frac{-4+8}{4}=1\\ x_2&=&~\frac{-4-8}{4}=-3 \end{array}$

For $0.2x^2+1.2x+1=0$, first multiply both sides of the equation by $5$ to get $x^2+6x+5=0$, and thus $a=1$, $b=6$ and $c=5$. This simplifies the following calculation:

$\begin{array}{rcl} x&=&~\frac{-6\pm\sqrt{6^2-4(1)(5)}}{2(1)}\\ &=&~\frac{-6\pm\sqrt{36-20}}{2}\\ &=&~\frac{-6\pm\sqrt{16}}{2}\\ x_1&=&~\frac{-6+4}{2}=-1\\ x_2&=&~\frac{-6-4}{2}=-5 \end{array}$

video image

What are Quadratic Functions?

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Graphing Quadratic Functions

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FOILing and Explanation for FOIL

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Solving Quadratic Equations by Taking Square Roots

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Solving Quadratic Equations by Factoring

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Factoring with Grouping

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Solving Quadratic Equations Using the Quadratic Formula

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Solving Quadratic Equations by Completing the Square

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Finding the Value that Completes the Square

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Using and Understanding the Discriminant

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Mathematics

Quadratic equations - word problems with solutions.

In this tutorial, you will learn:

  • Solving word problems with quadratic equations.
  • Interactive practice with randomly generated quadratic equations
  • How to build up a quadratic equation from a real life example.
  • How to solve the quadratic equation to find the required outcome.
  • How each question evolves to give you a perfect understanding in using quadratic equations for real world problems
  • Modelling real life situations with quadratic functions - mathematical modelling
  • Infinite square root problem

In order to learn how to solve quadratic equations by four different methods, please follow this tutorial ; it's detailed and offers plenty of worked examples and practice questions.

In this tutorial, I assume you have already got a good skill in solving quadratic equations as explained in the above tutorial. This tutorial primarily focuses on solving real-world problems involving quadratic equations.

If you think you need a bit more practice before dealing with word problems, here is a question generator and a programme to check your answers:

The Equation Generator

The following programme is interactive: by clicking on the buttons, you can generate a random equation and its solutions: they are randomly generated - and unlimited in number.

Quadratic Equation Generator

The quadratic solver.

A quadratic equation takes the form of ax 2 + bx + c where a and b are two integers, known as coefficients of x 2 and x respectively and c , a constant.

Enter a, b and c to find the solutions of the equations.

x 2 - x - 6 = 0, where a = 1; b=-1; c=-6

Quadratic Word Problems

The sum of two numbers is 27 and their product is 50. Find the numbers. Let one number be x. Then the other number is 50/x. x + 50/x = 27 X => x x 2 + 50 = 27x - 27x => x 2 - 27x + 50 = 0 (x -25)(x -2) = 0 (x -25) = 0 or (x -2) = 0 x = 25 or x = 2.

The length of a rectangle is 5 cm more than its width and the area is 50cm 2 . Find the length, width and the perimeter. Let the width be x. Then the length = x + 5. x(x + 5) = 50 x 2 + 5x = 50 -50 => x 2 + 5x - 50 = 0 (x + 10)(x -5) =0 (x + 10) = 0 or (x -5) =0 x = -10 or x = 5 - x = -10 is impossible to be a width Width = 5cm; so, the length = 10cm. Perimeter = 30cm.

The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area, if the longest side is 5. The hypotenuse = 5 x 2 + (x+1) 2 = 5 2 (Pythagoras' Theorem) x 2 + x 2 + 2x + 1 = 25 -25 => x 2 + x 2 + 2x - 24 = 0 2x 2 + 2x - 24 = 0 x 2 + x - 12 = 0 (x - 3)(x + 4) = 0 (x + 4) = 0 or (x - 3) = 0 x = -4 or x = 3 x = 3; Area = 1/2 x 3 x 4 = 6cm 2

The product of two numbers is 24 and the mean is 5. Find the numbers. Let one number = x; then the other = 24/x (x + 24/x)/2 = 5 X 2 => x + 24/x = 10 X x => x 2 + 24 = 10x - 10x => x 2 + -10x + 24 = 0 (x - 6)(x -4) = 0 (x - 6) = 0 or (x -4) = 0 x = 6 or x =4 The numbers are 6 or 4.

The sum of numbers is 9. The squares of the numbers is 41. Find the numbers. These are quadratic simultaneous equations. let the numbers be x and y. x + y = 9 x 2 + y 2 = 41 From the first equation, y = (9-x) Now substitute this in the second equation. x 2 + (9-x) 2 = 41 x 2 + 81 - 18x + x 2 = 41 2x 2 - 16x + 81 = 41 2x 2 - 16x + 40 = 0 x 2 - 8x + 20 = 0 (x - 5)(x -4) =0 (x - 5) = 0 or (x -4) =0 x = 5 or x = 4 Substitute in the first equation, y = 5 or 4 The numbers are 5 and 4.

Mathematical Modelling with Quadratic Functions

There are quite a few real life situations that can be modelled by a quadratic function in an accurate way. For instance, the motion of a ball, thrown upwards to move under gravity, can be easily modelled by a quadratic equation. Using the model, we can calculate the height of the ball if the time is known or vice versa.

A ball is thrown upwards from a rooftop, 80m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time t is h, which is given by, h = -16t 2 + 64t + 80.

  • What is the height reached by the ball after 1 second?
  • What is the maximum height reached by the ball?
  • How long will it take before hitting the ground?

Follow the graph along with the calculation for a better understanding:

moving ball

1) h = -16t 2 + 64t + 80 h = -16* 1*1 + 64*1 + 80 = 128m 2) Rearrange by the completing the square, we get: h = -16[t 2 - 4t - 5] h = -16[(t - 2) 2 - 9] h = -16(t - 2) 2 + 144 When the height is maximum, t = 2; therefore, maximum height = 144m. 3) When the ball hits the ground, h = 0; -16t 2 + 64t + 80 = 0 Divide the equation by -16 t 2 - 4t - 5 = 0 (t - 5)(t + 1) = 0 t = 5 or t = -1 The time cannot be negative; so, the time = 5 seconds.

A farmer wants to make a rectangular pen for his sheep. He has 60m fencing material to cover three sides with the other side being a brick wall. How should he use the fencing material to maximize the space for his sheep? How should he choose length and width of the pen to achieve his objective?

He just has to cover three sides; let the width be x. Then the length = (60-2x) Area of the pen = x(60-2x) = 60x -2x 2 Now let's sketch a graph for the quadratic equation. Which is as follows:

As you can see, the curve peaks at x = 15; when the width = 15m, the area is maximum. A very useful way to use quadratic equations in real life , indeed!

math word problem involving quadratic equation

Two resistors, when connected in series, have a total resistance of 25 Ohms. If they are connected in parallel, the value goes down to 6 Ohms. Find the values.

When they are in series, if one resistor is x, then the other is 25-x When they are in parallel, 1/6 = 1/x + 1/(25-x) 1/6 = 25/[x(25-x)] = 25/[25x - x 2 ] 25x - x 2 = 150 If ax 2 + bx + c = 0, then x = [-b ±√(b 2 - 4ac) ]/ 2a x 2 - 25x + 150 = 0 a = 1; b = -25; c = 150 x = -(-25) ±√((-25) 2 - 4(1)(150)) / 2(1) x = 25 ±√(625 - 600) / 2 x = 25 ±√(25) / 2 x = (25 ± 5 )/ 2 x = 15 or x = 10 So, the resistors are 15 Ohms or 10 Ohms.

The following picture shows the shape of a certain grass patch. If the area of the patch is 80m 2 , find k.

quadratic curve for grass patch

The total area = 5k + k(2k+1) = 5k + 2k 2 + k = 2k 2 + 6k Since the area is 80m 2 2k 2 + 6k = 80 2k 2 + 6k - 80 = 0 (2k - 10)(k + 8) = 0 k = 5 or k = -8 Since the length cannot be negative, k = 5.

The following picture shows the shape of a rectangle from which a smaller rectangular part is removed. If the remaining area of the larger rectangle is 35cm 2 , find k.

rectangular cardboard

The remaining area = k(2k+6) - 3k = 2k 2 + 6k -3k = 2k 2 + 3k Since the area is 35cm 2 2k 2 + 3k = 35 2k 2 + 3k - 35 = 0 (2k -7)(k + 5) = 0 k = 3.5 or k = -5 Since the length cannot be negative, k = 3.5.

The shortest side of a right-angled triangle is 6cm shorter than its hypotenuse. The difference in length of other two sides is 3cm. If the shortest side is n-3 , show that 2n 2 = 12n. Hence, find n.

pythagoras and quadratics

If the length of the shortest side is n-3 , the length of the hypotenuse and the other side are n+3 and n respectively. So, using Pythagoras Theorem, (n-3) 2 + n 2 = (n+3) 2 n 2 - 6n + 9 + n 2 = n 2 + 6n + 9 n 2 = 12n n 2 - 12n = 0 n(n - 12) = 0 n = 0 or n = 12. Since the length cannot be zero, n = 12.

Two cyclists move away from a town along two perpendicular paths at 20 mph and 40 mph respectively. The second cyclist starts the journey an hour later than the first one. Find the time taken for them to be 100 miles apart.

two cyclists

Suppose the time taken by the first cyclist is t; then the time taken by the other cyclist = (t-1) The distances travelled by them are 20t and 40(t-1) respectively. Using Pythagoras Theorem, (20t) 2 + [40(t-1)] 2 = 100 2 400t 2 + 1600(t-1) 2 = 10000 t 2 + 4t 2 - 8t + 4 = 25 5t 2 - 8t -21 = 0 (5t + 7)(t -3) = 0 t = 3 or t = -1.4 Since time cannot be negative, t = 3hrs.

Modelling Stopping Distance of a Vehicle with Speed

With the available data, the stopping distance - sum of thinking distance and breaking distance - can be modelled by a quadratic function. It is as follows:

stopping distance

If the stopping distance = d and the speed = v, d = 0.044v 2 + 1.1v

The following animation shows how the stopping distance, modelled by the above equation, changes with the speed:

Infinite Square Root Problem

Solve the following infinite square root problem:

infinite square root problem

20 + x = x 2 x 2 - x - 20 = 0 x = (1 ± √1 + 80)/2 x = (1 ± √81)/2 x = (1 ± 9)/2 x = 5 or x = -4 Since the square root of a negative number cannot be found, x = 5. So, the value of the above infinite square problem is 5.

This is a book designed for IGCSE, the international GCSE examination in the UK and the Commonwealth countries - in the age group 14 - 16. The questions, however, are superb for anyone who want to master mathematical concepts in this age group, regardless of the examinations that they sit for.

math word problem involving quadratic equation

The book contains lots of exercises, graded progressively from simple to harder, covering almost all the topics. If they are practised in the order given, the confidence of the students are naturally boosted to such an extent the need for any more resources becomes non-existent. There are two books in this series - book one and book two. I strongly recommend both for students. Please click the image, which will take you to Amazon book store.

🏆 Challenging Questions

1) Solve the following equation and find all the possible values of x: (x² - x - 5) x² - 2x - 3 = 1 2) Show that x 2 -2xy + 2y 2 can never be a negative.

Ad: The author of this site offers fully interactive tutorial on differentiation

differentiation course - teachable.com

Now, in order to complement what you have just learnt, work out the following questions:

  • The sum of squares of two consecutive even numbers is 244. Find the numbers.
  • The base length of a triangle is 2cm more than its height. The area is 24cm 2 . Find the length of hypotenuse and the perimeter of the triangle.
  • The length of a square is increased by a 5 th so that its new area is 44cm 2 more than the original value. Find the difference in perimeter of two shapes.
  • The length and width of a rectangular garden are 150m and 120m. A foot path of regular width is added to the boundary of the garden and the total area of the garden becomes 2800m 2 more than its original area. Find the width of the footpath.
  • Adam is about to embark on a journey on a narrow country lane that covers 32km and decides to go at x km/h. On second thoughts, he calculates that if he increases the speed by 4km/h, his journey time can be cut down by 4 hrs. Find x.
  • The reciprocal of the sum of reciprocals of two numbers is 6. The sum of numbers is 25. Find the numbers.
  • The speed of an ant is (2t + 10) and after travelling for t minutes, it covers a distance of 12m. Find t.
  • Two chords and a diameter form a triangle inside a circle. The radius is 5cm and one chord is 2cm longer than the other one. Find the perimeter and the area of the triangle.
  • The sum of a number and it reciprocal is 26/5. Find the number.
  • The product of two numbers is 20. The sum of squares is 41. Find the numbers.
  • The dimensions of the glass plate of a wedding photo are 18cm and 12cm respectively. A new rectangular frame of equal width all, the way round the photo, is about to be fitted around the it so that the area of the frame is the same as that of the glass. Find the width of the frame.
  • A group of acquaints went to a restaurants for a meal. When the bill for £175 was brought by a waiter, two of the cheeky ones from the group just sneaked off before the bill was paid, which resulted in the payment of extra £10 by each remaining individual. How many were in the group at first?
  • Ashwin and Donald decided to set out from two towns on their bikes, which are 247 miles apart, connected by a straight t Roman road in England. When they finally met up somewhere between the two towns, Ashwin had been cycling for 9 miles a day. The number of days for the whole adventure is 3 more than the number of miles that Donald had been cycling in a day. How many miles did each cycle?
  • When a two-digit number is divided by the product of the two digits, the answer is 2 and if 27 is added to the number, the original number turns into a new number with the digits being swapped around. Find the number.
  • There are three numbers: the difference of the differences of them is 5. The sum and product are 44 and 1950 respectively. Find the numbers.
  • Find the two numbers, whose sum is 19 and the product of the difference and the greater, is 60.
  • A boy was asked his age: "If you add the square root of it to half of it, and then subtract 12, the answer will be nothing," replied the boy. What was his age?
  • A group of army cadets, consisting of 1066 men, form two squares in front of a garrison. In the side of one square, there are 4 more men than the other. How many men are in each side of the squares?
  • The height of a triangle is 4cm less than three times its base length. If the area is 80 cm 2 , find the lengths of the base and height.
  • An isosceles triangle is inscribed in a circle in such a way that its longest side, which goes through the centre is √50cm. Find the area of the triangle.
  • A butcher bought some pheasants for £100.00. Had each cost £1 less, he would have bought 5 more. How many pheasants did he buy?
  • The sum of reciprocals of two consecutive integers is 13/42. Find the integers.
  • The product of two numbers and the difference between them are 289 and 20 respectively. Find the numbers.
  • The sum of two numbers is 20. The sum of squares is 250. Find the numbers.
  • The area of a square exceeds twice that of another by 56cm 2 . If the difference of the perimeter between the two is 24cm, find the area of the smaller square.

Move the mouse over, just below this, to see the answers:

  • 24cm, 24cm 2
  • 117 and 130 miles

Do you need detailed solutions to the above questions? Please click the following and I will send the answers:

Detailed Solutions for the above Quadratic Questions As a PDF for Just £3.99

Now that you have read this tutorial, you will find the following tutorials very helpful too:

  • Solving Quadratic Equations - interactive
  • Transformations of Graphs - interactive
  • Solving Simultaneous Equations
  • Simultaneous Equations - word problems
  • Iteration - interactive
  • Factorisation

Amazon Publications

Ace the Toughest GCSE Maths Topics (9-1) with Clear Worked Examples! This comprehensive guide breaks down even the most challenging concepts into step-by-step explanations and provides loads of practice problems to help you master the GCSE Maths (9-1) curriculum.

Unlock the secrets of logarithms and exponential functions with this book. Our comprehensive guide is packed with clear explanations and worked examples to help you master these powerful mathematical tools.

Master Binomial Theorems with Ease! This comprehensive guide dives deep into binomial theorems, providing clear explanations, step-by-step examples, and powerful practice problems to solidify your understanding. Boost your math skills and ace your exams with this essential resource!

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Mathematics LibreTexts

6.7: Applications Involving Quadratic Equations

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  • Page ID 23744

Learning Objectives

  • Set up and solve applications involving relationships between real numbers.
  • Set up and solve applications involving geometric relationships involving area and the Pythagorean theorem.
  • Set up and solve applications involving the height of projectiles.

Number Problems

The algebraic setups of the word problems that we have previously encountered led to linear equations. When we translate the applications to algebraic setups in this section, the setups lead to quadratic equations. Just as before, we want to avoid relying on the “guess and check” method for solving applications. Using algebra to solve problems simplifies the process and is more reliable.

Example \(\PageIndex{1}\)

One integer is \(4\) less than twice another integer, and their product is \(96\). Set up an algebraic equation and solve it to find the two integers.

First, identify the variables. Avoid two variables by using the relationship between the two unknowns.

Screenshot (342).png

The key phrase, “their product is \(96\),” indicates that we should multiply and set the product equal to \(96\).

\(n\cdot (2n-4)=96\)

Once we have the problem translated to a mathematical equation, we then solve. In this case, we can solve by factoring. The first step is to write the equation in standard form:

\(\begin{array} {cc} {n\cdot (2n-4)=96}&{\color{Cerulean}{Distribute\:n.}}\\{2n^{2}-4n=96}&{\color{Cerulean}{Subtract\:96\:from\:both\:sides.}}\\{2n^{2}-4n-96=0}&{} \end{array}\)

Next, factor completely and set each variable factor equal to zero.

\(\begin{array}{cc}{2n^{2}-4n-96=0}&{\color{Cerulean}{Factor\:out\:the\:GCF,\:2.}}\\{2(n^{2}-2n-48)=0}&{\color{Cerulean}{Factor\:the\:resulting\:trinomial.}}\\{2(n+6)(n-8)=0}&{\color{Cerulean}{Set\:each\:variable\:factor\:equal\:to\:zero.}} \end{array}\)

\(\begin{array}{ccc}{n+6=0}&{\text{or}}&{n-8=0}\\{n=-6}&{}&{n=8} \end{array}\)

The problem calls for two integers whose product is \(+96\). The product of two positive numbers is positive and the product of two negative numbers is positive. Hence we can have two sets of solutions. Use \(2n−4\) to determine the other integers.

\(\begin{array} {cc} {n=-6}&{n=8}\\{2n-4=2(\color{OliveGreen}{-6}\color{black}{)-4}}&{2n-4=2(\color{OliveGreen}{8}\color{black}{)-4}}\\{=-12-4}&{=16-4}\\{=-16}&{=12} \end{array}\)

Two sets of integers solve this problem: {\(8, 12\)} and {\(−6, −16\)}. Notice that \((8)(12) = 96\) and \((−6)(−16) = 96\); our solutions check out.

With quadratic equations, we often obtain two solutions for the identified unknown. Although it may be the case that both are solutions to the equation, they may not be solutions to the problem. If a solution does not solve the original application, then we disregard it.

Recall that consecutive odd and even integers both are separated by two units.

Screenshot (343).png

Example \(\PageIndex{2}\)

  • The product of two consecutive positive odd integers is \(99\). Find the integers.

Let \(n\) represent the first positive odd integer.

Let \(\color{OliveGreen}{n+2}\) represent the next positive odd integer.

The key phase, “product…is 99,” indicates that we should multiply and set the product equal to \(99\).

\(n\cdot (n+2)=99\)

Rewrite the quadratic equation in standard form and solve by factoring.

\(\begin{aligned} n^{2}+2n&=99 \\ n^{2}+2n-99&=0 \\ (n-9)(n+11)&=0 \end{aligned}\)

\(\begin{array}{ccc}{n-9=0}&{\text{or}}&{n+11=0}\\{n=9}&{}&{n=-11} \end{array}\)

Because the problem asks for positive integers, \(n=9\) is the only solution. Back substitute to determine the next odd integer.

\(\begin{aligned} n+2&=\color{OliveGreen}{9}\color{black}{+2} \\ &=11 \end{aligned}\)

The consecutive positive odd integers are \(9\) and \(11\).

Example \(\PageIndex{3}\)

Given two consecutive positive odd integers, the product of the larger and twice the smaller is equal to \(70\). Find the integers.

Let \(n\) represent the smaller positive odd integer.

Let \(n+2\) represent the next positive odd integer.

The key phrase “twice the smaller” can be translated to \(2n\). The phrase “product…is 70” indicates that we should multiply this by the larger odd integer and set the product equal to \(70\).

\((n+2)\cdot 2n=70\)

Solve by factoring.

\(\begin{array}{cc}{(n+2)\cdot 2n=70}&{\color{Cerulean}{Distribute.}}\\{2n^{2}+4n=70}&{\color{Cerulean}{Subtract\:70\:from\:both\:sides.}}\\{2n^{2}+4n-70=0}&{\color{Cerulean}{Factor\:out\:the\:GCF,\:2.}}\\{2(n^{2}+2n-35)=0}&{\color{Cerulean}{Factor\:the\:resulting\:trinomial.}}\\{2(n-5)(n+7)=0}&{\color{Cerulean}{Set\:each\:variable\:factor\:equal\:to\:zero.}} \end{array}\)

\(\begin{array}{ccc}{n-5=0}&{\text{or}}&{n+7=0}\\{n=5}&{}&{n=-7} \end{array}\)

Because the problem asks for positive integers, \(n=5\) is the only solution.

Back substitute into \(n + 2\) to determine the next odd integer.

\(\begin{aligned} n+2&=\color{OliveGreen}{5}\color{black}{+2}\\&=7 \end{aligned}\)

The positive odd integers are \(5\) and \(7\).

Exercise \(\PageIndex{1}\)

The product of two consecutive positive even integers is \(168\). Find the integers.

The positive even integers are \(12\) and \(14\).

Geometry Problems

When working with geometry problems, it is helpful to draw a picture. Below are some area formulas that you are expected to know. (Recall that \(π≈3.14\).)

Example \(\PageIndex{4}\)

The floor of a rectangular room has a length that is \(4\) feet more than twice its width. If the total area of the floor is \(240\) square feet, then find the dimensions of the floor.

Screenshot (344).png

Use the formula \(A=l⋅w\) and the fact that the area is \(240\) square feet to set up an algebraic equation.

\(\begin{aligned} A&=l\cdot w \\ \color{OliveGreen}{240}&\color{black}{=(}\color{OliveGreen}{2w+4}\color{black}{)\cdot w} \end{aligned}\)

\(\begin{array} {ccc} {w-10=0}&{\text{or}}&{w+12=0} \\{w=10}&{}&{w=-12} \end{array}\)

At this point we have two possibilities for the width of the rectangle. However, since a negative width is not defined, choose the positive solution, \(w=10\). Back substitute to find the length.

\(\begin{aligned} 2w+4&=2(\color{OliveGreen}{10}\color{black}{)+4} \\ &=20+4 \\ &=24 \end{aligned}\)

The width is \(10\) feet and the length is \(24\) feet.

It is important to include the correct units in the final presentation of the answer. In the previous example, it would not make much sense to say the width is \(10\). Make sure to indicate that the width is \(10\) feet.

Example \(\PageIndex{5}\)

The height of a triangle is \(3\) inches less than twice the length of its base. If the total area of the triangle is \(7\) square inches, then find the lengths of the base and height.

Screenshot (345).png

Use the formula \(A=\frac{1}{2}bh\) and the fact that the area is \(7\) square inches to set up an algebraic equation.

\(\begin{aligned} A&=\frac{1}{2} b\cdot h \\ \color{OliveGreen}{7}&\color{black}{=\frac{1}{2}b(\color{OliveGreen}{2b-3}\color{black}{)}} \end{aligned}\)

To avoid fractional coefficients, multiply both sides by \(2\) and then rewrite the quadratic equation in standard form.

Factor and then set each factor equal to zero.

\(\begin{array}{ccc}{2b-7=0}&{\text{or}}&{b+2=0}\\{2b=7}&{}&{b=-2}\\{b=\frac{7}{2}}&{}&{} \end{array}\)

In this case, disregard the negative answer; the length of the base is \(\frac{7}{2}\) inches long. Use \(2b−3\) to determine the height of the triangle.

The base measures \(\frac{7}{2} = 3\frac{1}{2}\) inches and the height is \(4\) inches.

The base of a triangle is \(5\) units less than twice the height. If the area is \(75\) square units, then what is the length of the base and height?

The height is \(10\) units and the base is \(15\) units.

Recall that a right triangle is a triangle where one of the angles measures \(90\)°. The side opposite of the right angle is the longest side of the triangle and is called the hypotenuse. The Pythagorean theorem gives us a relationship between the legs and hypotenuse of any right triangle, where \(a\) and \(b\) are the lengths of the legs and \(c\) is the length of the hypotenuse:

Screenshot (346).png

Given certain relationships, we use this theorem when determining the lengths of sides of right triangles.

Example \(\PageIndex{6}\)

The hypotenuse of a right triangle is \(10\) inches. If the short leg is \(2\) inches less than the long leg, then find the lengths of the legs.

Screenshot (347).png

Given that the hypotenuse measures \(10\) inches, substitute its value into the Pythagorean theorem and obtain a quadratic equation in terms of \(x\).

\(\begin{aligned} a^{2}+b^{2}&=c^{2} \\ (\color{OliveGreen}{x-2}\color{black}{)^{2}+}\color{OliveGreen}{x}\color{black}{^{2}}&=\color{OliveGreen}{10}\color{black}{^{2}} \end{aligned}\)

Multiply and rewrite the equation in standard form.

\(\begin{aligned} (x-2)^{2}+x^{2}&=10^{2} \\ x^{2}-4x+4+x^{2}&=100 \\ 2x^{2}-4x-96&=0 \end{aligned}\)

Once it is in standard form, factor and set each variable factor equal to zero.

\(\begin{aligned} 2x^{2}-4x-96&=0\\ 2(x^{2}-2x-48)&=0 \\ 2(x+6)(x-8)&=0 \end{aligned}\)

\(\begin{array}{ccc}{x+6=0}&{\text{or}}&{x-8=0}\\{x=-6}&{}&{x=8} \end{array}\)

Because lengths cannot be negative, disregard the negative answer. In this case, the long leg measures \(8\) inches. Use \(x−2\) to determine the length of the short leg.

\(\begin{aligned} x-2&=\color{OliveGreen}{8}\color{black}{-2} \\ &=6 \end{aligned}\)

The short leg measures \(6\) inches and the long leg measures \(8\) inches.

Example \(\PageIndex{7}\)

One leg of a right triangle measures \(3\) centimeters. The hypotenuse of the right triangle measures \(3\) centimeters less than twice the length of the unknown leg. Find the measure of all the sides of the triangle.

Screenshot (348).png

To set up an algebraic equation, we use the Pythagorean theorem.

\(\begin{aligned} a^{2}+b^{2}&=c^{2} \\ \color{OliveGreen}{3}\color{black}{^{2}+}\color{OliveGreen}{x}\color{black}{^{2}}&=(\color{OliveGreen}{2x-3}\color{black}{)^{2}} \end{aligned}\)

\(\begin{aligned} 3^{2}+x^{2}&=(2x-3)^{2} \\ 9+x^{2}&=4x^{2}-12x+9 \\ 0&=3x^{2}-12x \\ 0&=3x(x-4) \end{aligned}\)

\(\begin{array}{ccc}{3x=0}&{\text{or}}&{x-4=0}\\{x=0}&{}&{x=4} \end{array}\)

Disregard \(0\). The length of the unknown leg is \(4\) centimeters. Use \(2x−3\) to determine the length of the hypotenuse.

The sides of the triangle measure \(3\) centimeters, \(4\) centimeters, and \(5\) centimeters.

Exercise \(\PageIndex{2}\)

The hypotenuse of a right triangle measures \(13\) units. If one leg is \(2\) units more than twice that of the other, then find the length of each leg.

The two legs measure \(5\) units and \(12\) units.

Projectile Problems

The height of an object launched upward, ignoring the effects of air resistance, can be modeled with the following formula:

\[\text{height}=-\frac{1}{2}gt^{2}+v_{0}t+s_{0}\]

Using function notation, which is more appropriate, we have

\[h(t)=-\frac{1}{2}gt^{2}+v_{0}t+s_{0}\]

With this formula, the height can be calculated at any given time \(t\) after the object is launched. The coefficients represent the following:

We consider only problems where the acceleration due to gravity can be expressed as \(g=32\) ft/sec\(^{2}\). Therefore, in this section time will be measured in seconds and the height in feet. Certainly though, the formula is valid using units other than these.

Example \(\PageIndex{8}\)

The height of a projectile launched upward at a speed of \(32\) feet/second from a height of \(128\) feet is given by the function \(h(t)=−16t^{2}+32t+128\). How long does it take to hit the ground?

An inefficient method for finding the time to hit the ground is to simply start guessing at times and evaluating. To do this, construct a chart.

Screenshot (349).png

Use the table to sketch the height of the projectile over time.

Screenshot (350).png

We see that at \(4\) seconds, the projectile hits the ground. Note that when this occurs, the height is equal to \(0\). Now we need to solve this problem algebraically. To find the solution algebraically, use the fact that the height is \(0\) when the object hits the ground. We need to find the time, \(t\), when \(h(t)=0\).

\(h(t)=-16t^{2}+32t+128 \\ \color{Cerulean}{\downarrow} \\ 0 =-16t^{2}+32t+128\)

Solve the equation by factoring

\(\begin{aligned} 0 &=-16t^{2}+32t+128 \\ 0 &=-16(t^{2}-2t-8) \\ 0&=-16(t-4)(t+2) \end{aligned}\)

Now set each variable factor to zero.

\(\begin{array}{ccc}{t-4=0}&{\text{or}}&{t+2=0}\\{t=4}&{}&{t=-2} \end{array}\)

As expected, the projectile hits the ground at \(t=4\) seconds. Disregard \(−2\) as a solution because negative time is not defined.

The projectile hits the ground \(4\) seconds after it is launched.

Example \(\PageIndex{9}\)

The height of a certain book dropped from the top of a \(144\) foot building is given by \(h(t)=−16t^{2}+144\). How long does it take to hit the ground?

Find the time \(t\) when the height \(h(t)=0\).

\(\begin{aligned}0&=-16t^{2}+144 \\ 0&=-16(t^{2}-9) \\ 0&=-16(t+3)(t-3) \end{aligned}\)

\(\begin{array}{ccc}{t+3=0}&{\text{or}}&{t-3=0}\\{t=-3}&{}&{t=3} \end{array}\)

The book takes \(3\) seconds to hit the ground when dropped from the top of a \(144\)-foot building.

Exercise \(\PageIndex{3}\)

The height of a projectile, shot straight up into the air from the ground, is given by \(h(t)=−16t^{2}+80t\). How long does it take to come back down to the ground?

It will take 5 seconds to come back down to the ground.

Key Takeaways

  • It is best to translate a word problem to a mathematical setup and then solve using algebra. Avoid using the “guess and check” method of solving applications in this section.
  • When solving applications, check that your solutions make sense in the context of the question. For example, if you wish to find the length of the base of a triangle, then you would disregard any negative solutions.
  • It is important to identify each variable and state in a sentence what each variable represents. It is often helpful to draw a picture.

Exercise \(\PageIndex{4}\) Number Problems

Set up an algebraic equation and then solve.

  • One integer is five times another. If the product of the two integers is \(80\), then find the integers.
  • One integer is four times another. If the product of the two integers is \(36\), then find the integers.
  • An integer is one more than four times another. If the product of the two integers is \(39\), then find the integers.
  • An integer is \(3\) more than another. If the product of the two integers is \(130\), then find the integers.
  • An integer is \(2\) less than twice another. If the product of the two integers is \(220\), then find the integers.
  • An integer is \(3\) more than twice another. If the product of the two integers is \(90\), then find the integers.
  • One integer is \(2\) units more than another. If the product of the two integers is equal to five times the larger, then find the two integers.
  • A positive integer is \(1\) less than twice another. If the product of the two integers is equal to fifteen times the smaller, then find the two integers.
  • A positive integer is \(3\) more than twice a smaller positive integer. If the product of the two integers is equal to six times the larger, then find the integers.
  • One positive integer is \(3\) more than another. If the product of the two integers is equal to twelve times the smaller, then find the integers.
  • An integer is \(3\) more than another. If the product of the two integers is equal to \(2\) more than four times their sum, then find the integers.
  • An integer is \(5\) more than another. If the product of the two integers is equal to \(2\) more than twice their sum, then find the integers.
  • The product of two consecutive positive even integers is \(120\). Find the integers.
  • The product of two consecutive positive integers is \(110\). Find the integers.
  • The product of two consecutive positive integers is \(42\). Find the integers.
  • The product of two consecutive positive odd integers is equal to \(1\) less than seven times the sum of the integers. Find the integers.
  • The product of two consecutive positive even integers is equal to \(22\) more than eleven times the sum of the integers. Find the integers.
  • The sum of the squares of two consecutive positive odd integers is \(74\). Find the integers.
  • The sum of the squares of two consecutive positive even integers is \(100\). Find the integers.
  • The sum of the squares of two consecutive positive integers is \(265\). Find the integers.
  • The sum of the squares of two consecutive positive integers is \(181\). Find the integers.
  • For two consecutive positive odd integers, the product of twice the smaller and the larger is \(126\). Find the integers.
  • For two consecutive positive even integers, the product of the smaller and twice the larger is \(160\). Find the integers.

1. {\(4, 20\)} or {\(−4, −20\)}

3. \(3, 13\)

5. {\(11, 20\)} or {\(−22, −10\)}

7. {\(5, 7\)} or {\(−2, 0\)}

9. \(6, 15\)

11. {\(7, 10\)} or {\(−2, 1\)}

13. \(10, 12\)

15. \(10, 11\)

17. \(13, 15\)

19. \(5, 7\)

21. \(11, 12\)

23. \(7, 9\)

Exercise \(\PageIndex{5}\) Geometry Problems

  • The width of a rectangle is \(7\) feet less than its length. If the area of the rectangle is \(170\) square feet, then find the length and width.
  • The length of a rectangle is \(2\) feet more than its width. If the area of the rectangle is \(48\) square feet, then find the length and width.
  • The width of a rectangle is \(3\) units less than the length. If the area is \(70\) square units, then find the dimensions of the rectangle.
  • The width of a rectangle measures one half of the length. If the area is \(72\) square feet, then find the dimensions of the rectangle.
  • The length of a rectangle is twice that of its width. If the area of the rectangle is \(72\) square inches, then find the length and width.
  • The length of a rectangle is three times that of its width. If the area of the rectangle is \(75\) square centimeters, then find the length and width.
  • The length of a rectangle is \(2\) inches more than its width. The area of the rectangle is equal to \(12\) inches more than three times the perimeter. Find the length and width of the rectangle.
  • The length of a rectangle is \(3\) meters more than twice the width. The area of the rectangle is equal to \(10\) meters less than three times the perimeter. Find the length and width of the rectangle.
  • A uniform border is to be placed around an \(8\)-inch-by-\(10\)-inch picture. If the total area including the border must be \(224\) square inches, then how wide should the border be?

Screenshot (351).png

10. A \(2\)-foot brick border is constructed around a square cement slab. If the total area, including the border, is \(121\) square feet, then what are the dimensions of the slab?

11. The area of a picture frame including a \(2\)-inch wide border is \(99\) square inches. If the width of the inner area is \(2\) inches more than its length, then find the dimensions of the inner area.

Screenshot (352).png

12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of \(2\) inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be \(50\) cubic inches?

Screenshot (353).png

13. The height of a triangle is \(3\) inches more than the length of its base. If the area of the triangle is \(44\) square inches, then find the length of its base and height.

14. The height of a triangle is \(4\) units less than the length of the base. If the area of the triangle is \(48\) square units, then find the length of its base and height.

15. The base of a triangle is twice that of its height. If the area is \(36\) square centimeters, then find the length of its base and height.

16. The height of a triangle is three times the length of its base. If the area is \(73\frac{1}{2}\) square feet, then find the length of the base and height.

17. The height of a triangle is \(1\) unit more than the length of its base. If the area is \(5\) units more than four times the height, then find the length of the base and height of the triangle.

18. The base of a triangle is \(4\) times that of its height. If the area is \(3\) units more than five times the height, then find the length of the base and height of the triangle.

19. The diagonal of a rectangle measures \(5\) inches. If the length is \(1\) inch more than its width, then find the dimensions of the rectangle.

20. The diagonal of a rectangle measures \(10\) inches. If the width is \(2\) inches less than the length, then find the area of the rectangle.

21. If the sides of a right triangle are consecutive even integers, then what are their measures?

22. The hypotenuse of a right triangle is \(13\) units. If the length of one leg is \(2\) more than twice the other, then what are their lengths?

23. The shortest leg of a right triangle measures \(9\) centimeters and the hypotenuse measures \(3\) centimeters more than the longer leg. Find the length of the hypotenuse.

24. The long leg of a right triangle measures \(24\) centimeters and the hypotenuse measures \(4\) centimeters more three times the short leg. Find the length of the hypotenuse.

1. Length: \(17\) feet; width: \(10\) feet

3. Length: \(10\) units; width: \(7\) units

5. Length: \(12\) inches; width: \(6\) inches

7. Length: \(14\) inches; width: \(12\) inches

9. \(3\) inches

11. \(5\) inches by \(7\) inches

13. Base: \(8\) inches; height: \(11\) inches

15. Base: \(12\) centimeters; height: \(6\) centimeters

17. Base: \(9\) units; height: \(10\) units

19. \(3\) inches by \(4\) inches

21. \(6\) units, \(8\) units, and \(10\) units

23. \(15\) centimeters

Exercise \(\PageIndex{6}\) Projectile Problems

  • The height of a projectile launched upward at a speed of \(32\) feet/second from a height of \(48\) feet is given by the function \(h(t)=−16t^{2}+32t+48. How long will it take the projectile to hit the ground?
  • The height of a projectile launched upward at a speed of \(16\) feet/second from a height of \(192\) feet is given by the function \(h(t)=−16t^{2}+16t+192\). How long will it take to hit the ground?
  • An object launched upward at a speed of \(64\) feet/second from a height of \(80\) feet. How long will it take the projectile to hit the ground?
  • An object launched upward at a speed of \(128\) feet/second from a height of \(144\) feet. How long will it take the projectile to hit the ground?
  • The height of an object dropped from the top of a \(64\)-foot building is given by \(h(t)=−16t^{2}+64\). How long will it take the object to hit the ground?
  • The height of an object dropped from an airplane at \(1,600\) feet is given by \(h(t)=−16t^{2}+1,600\). How long will it take the object to hit the ground?
  • An object is dropped from a ladder at a height of \(16\) feet. How long will it take to hit the ground?
  • An object is dropped from a \(144\)-foot building. How long will it take to hit the ground?
  • The height of a projectile, shot straight up into the air from the ground at \(128\) feet/second, is given by \(h(t)=−16t^{2}+128t\). How long does it take to come back down to the ground?
  • A baseball, tossed up into the air from the ground at \(32\) feet/second, is given by \(h(t)=−16t^{2}+32t\). How long does it take to come back down to the ground?
  • How long will it take a baseball thrown into the air at \(48\) feet/second to come back down to the ground?
  • A football is kicked up into the air at \(80\) feet/second. Calculate how long will it hang in the air.

1. \(3\) seconds

3. \(5\) seconds

5. \(2\) seconds

7. \(1\) second

9. \(8\) seconds

11. \(3\) seconds

Exercise \(\PageIndex{7}\) Discussion Board Topics

  • Research and discuss the life of Pythagoras.
  • If the sides of a square are doubled, then by what factor is the area increased? Why?
  • Design your own geometry problem involving the area of a rectangle or triangle. Post the question and a complete solution on the discussion board.
  • Write down your strategy for setting up and solving word problems. Share your strategy on the discussion board.

1. Answers may vary

3. Answers may vary

math word problem involving quadratic equation

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Projectile Motion & Quadratic Equations

Projectiles General Word Problems Max/Min Problems

Calculus involves lots of finding maximums, minimums, and zeroes. You will get a taste of that with quadratic word problems. In fact, if/when you reach calculus, you will discover that some of the homework exercises will be identical to those you're doing now; it's just that you'll have new tools for finding the answers.

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A major category of quadratic-equation word problems relates to what is called projectile motion. For our purposes, a projectile is any object that is thrown, shot, or dropped. Almost always, in this context, the object is initially moving directly up or straight down. (If it starts by going up then, naturally, it will later be coming back down.) This initial movement speed is the velocity.

What does the velocity's sign say about the object?

The initial velocity of the object, in these exercises, tells us how the object was released. The initial value of the velocity will be either zero (so the object was just dropped), positive (so it was thrown or shot upward), or negative (so the object was thrown downward).

What does " g " stand for?

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In projectile-motion exercises, the coefficient on the squared term is −½  g . The g stands for the constant of gravity (on Earth), which is −9.8 meters per second square (that is meters per second per second) in metric terms, or −32 feet per second squared in Imperial terms. The "minus" signs reflect the fact that Earth's gravity pulls us, and the object in question, downward.

What does "per second squared" mean?

Acceleration (being the change in speed, rather than the speed itself) is measured in terms of how much the velocity changes per unit time. So, if the velocity of an object is measured in feet per second, then that object's acceleration says how much that velocity changes per unit time; that is, acceleration measures how much the feet per second changes per second. And this duplicate "per second" is how we get "second squared". It's from the physics of the situation.

Which value should I use for gravity?

If a projectile-motion exercise is stated in terms of feet, miles, or some other Imperial unit, then use −32 for gravity; if the units are meters, centimeters, or some other metric unit, then use −9.8 for gravity.

What is the projectile-motion equation?

The projectile-motion equation is s ( t ) = −½  g x 2 +  v 0 x  +  h 0 , where g is the constant of gravity, v 0 is the initial velocity (that is, the velocity at time t  = 0 ), and h 0 is the initial height of the object (that is, the height at of the object at t  = 0 , the time of release).

Yes, you'll need to keep track of all of this stuff when working with projectile motion.

  • An object is launched at 19.6 meters per second (m/s) from a 58.8 -meter tall platform. The equation for the object's height s at time t seconds after launch is s ( t ) = −4.9 t 2 + 19.6 t  + 58.8 , where s is in meters. When does the object strike the ground?

What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. So I'm looking for the time when the height is s  = 0 . I'll set s equal to zero, and solve:

0 = −4.9 t 2 + 19.6 t + 58.8 0 = t 2 − 4 t − 12 0 = ( t − 6)( t + 2)

Then t = 6 or t  = −2 . The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x -axis at −2 , but negative time won't work in this word problem.) So " t  = −2 " is an extraneous solution, and I'll ignore it.

Instead, my answer to this exercise (that is, the correct answer in context) is the other solution value. They asked me for the time, and time here is measured in seconds, so my answer is:

The object strikes the ground six seconds after launch.

Note the construction of the height equation in the problem above. (Yes, we went over this at the beginning, but you're really gonna need this info, so we're revisiting.)

The initial launch height was 58.8 meters, and the constant term was " 58.8 ". The initial velocity (or launch speed) was 19.6 m/s, and the coefficient on the linear term was " 19.6 ". This is always true for these up/down projectile motion problems. (If you have an exercise with sideways motion, the equation will have a different form, but they'll always give you that equation.) The initial velocity is the coefficient for the middle term, and the initial height is the constant term.

And the coefficient on the leading term comes from the force of gravity. This coefficient is negative, since gravity pulls downward, and the value will either be " −4.9 " (if your units are "meters") or " −16 " (if your units are "feet"). Yes, these values are half of the values listed for the gravity constant at the beginning of this page; they've had the ½ multiplied through.

In general, the projectile-motion equation's format is:

s ( t ) = − gt 2 + v 0 t + h 0

...where " g " here is the " 4.9 " or the " 16 " derived from the value of the force of gravity (technically, it's the force of gravity on Earth ), " v 0 " ("vee-naught", or "vee-sub-zero") is the initial velocity, and " h 0 " ("aitch-naught", or "aitch-sub-zero") is the initial height.

Memorize this equation (or at least its meaning), because you may need to know this on the test.

  • An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height?

Hmm... They didn't give me the equation this time. But that's okay, because I can create the equation from the information that they did give me. The initial height is 80 feet above ground and the initial speed is 64 ft/s upward. Since my units are feet, then the number for gravity will be −16 , and my equation is:

s ( t ) = −16 t 2 + 64 t + 80

They want me to find the maximum height. For a negative quadratic like this, the maximum will be at the vertex of the upside-down parabola. So they really want me to find the vertex. From graphing , I know how to find the vertex; in this case, the vertex is at ( h ,  k ) = (2, 144):

h = − b / 2 a = −(64)/2(−16) = −64/−32 = 2 k = s (2) = −16(2) 2 + 64(2) + 80 = −16(4) + 128 + 80 = 208 − 64 = 144

But what do the coordinates of this vertex tell me? According to my equation, I'm plugging in time values and extracting height values, so the input 2 was the time and the output 144 is the height.

It takes two seconds to reach the maximum height of 144 feet.

As long as you label clearly, you don't need a complete sentence (like I used above) for your hand-in answer. So you could also give you answer as, "time:  2  secs; height:  144  ft".

  • An object is launched from ground level directly upward at 39.2 m/s. For how long is the object at or above a height of 34.3 meters?

My units this time are "meters", so the gravity number will be −4.9 . Since the object started at ground level, the initial height was 0 . Then my equation is:

s ( t ) = −4.9 t 2 + 39.2 t

Since this is a negative quadratic, the graph is an upside-down parabola. I can find the two times when the object is exactly 34.3 meters high, and I know that the object will be above 34.3 meters the whole time in between.

Why "two times", and how do I know that the time period is between those two times? Because the first time will be when the object passes a height of 34.3 meters on its way up to its maximum height, and the second time when be when it passes 34.3 meters as it is falling back down to the ground. So I have to solve the following:

−4.9 t 2 + 39.2 t = 34.3 t 2 − 8 t + 7 = 0 ( t − 7)( t − 1) = 0

The two solutions are at times t  = 1 and t  = 7 . So the object is at 34.3  meters at one second after launch (going up) and againt at seven seconds after launch (coming back down). Subtracting to find the difference, I find that:

The object is at or above 34.3  meters for six seconds.

Again, you don't technically need a complete sentence for your hand-in answer; saying " 6 secs" is probably good enough. But definitely do include the unit "seconds" on your answer.

Don't be surprised if many of your exercises work out as "neatly" as the above examples have. Many textbooks still engineer their exercises carefully, so that you can solve by factoring (that is, by quickly doing the algebra).

However (fair warning!), heavy dependence on calculators is leading more texts to create "interesting" (that is to say, needlessly complicated) exercises, so some (or all) of your exercises may involve much more messy computations than have been displayed here. If so, study these "neat" examples carefully, until you are quite sure you follow the reasoning.

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  • After the semester is over, you discover that the math department has changed textbooks (again) so the bookstore won't buy back your nearly-new book. You and your friend Herman decide to get creative. You go to the roof of a twelve-story building and look over the edge to the reflecting pool 160  feet below. You drop your book over the edge at the same instant that Herman chucks his book straight down at 48  feet per second. By how many seconds does his book beat yours into the water? (Round your answer to two decimal places.)

Our initial launch heights will be the same: we're both launching from 160  feet above ground. And the gravity number, since we're working in feet, will be −16 . My initial velocity is zero, since I just dropped my book, but my buddy Herman's velocity is a negative 48 , the negative coming from the fact that he chucked his book down rather than up. So our "height" equations are:

mine: s ( t ) = −16 t 2 + 160 his: s ( t ) = −16 t 2 − 48 t + 160

In each case, I need to find the time for the books to reach a height of zero (a height of zero being "ground level"), so:

mine: 0 = −16 t 2 + 160

t 2 − 10 = 0

his: 0 = −16 t 2 − 48 t + 160

t 2 + 3 t − 10 = 0

( t + 5)( t − 2) = 0

t = −5, +2

I will ignore the negative time values as being irrelevant in this context. (There may be a case, at some point, in some exercise, where a negative value does happen to be useful within a given context, so I do need to remember to think about the values and the context, rather than reflexively discarding any and all negative solutions.)

Herman's book hits the water about 1.16 seconds sooner than mine does.

Every once in a while, they'll get clever and put a "projectile" problem into a different environment. The equation will remain the same in structure, but you may have to account for a different value for gravity.

  • The International Space Agency has finally landed a robotic explorer on an extra-solar planet. Some probes are extended from the lander's body to conduct various tests. To demonstrate the crushing weight of gravity on this planet, the lander's camera is aimed at a probe's ground-level ejection port, and the port launches a baseball directly upwards at 147 feet per second (ft/s), about the top speed of a professional pitcher. The force due to gravity on this planet is 98  ft/s 2 . Assuming no winds and that the probe can scurry out of the way in time, how long will it take for the ball to smack back into the surface?

To set up my equation for this exercise, I need to keep in mind that the value of the coefficient g from the projectile-motion equation above is one-half of the value of the force due to gravity in a given locale.

In physics, there is the "universal gravitational constant " G , being the gravitational pull inherent to our universe (or at least our region of it). Every object exerts its own gravitational force, which is related to its own mass and the universal constant G . In the "projectile motion" formula, the " g " is half of the value of the gravitational force for that particular body. For instance, the gravitational force on Earth is a downward 32 ft/s 2 , but we used 16 in the equation.

So g for my equation this time will by 98 ÷ 2 = 49 feet per second squared. Then:

s = −49 t 2 + 147 t 0 = −49 t 2 + 147 t 0 = t 2 − 3 t 0 = t  ( t − 3)

Then t = 0 or t = 3 . The first solution represents when the ball was launched, so the second solution is the one I want.

It takes three seconds for the ball to hit the ground.

Note: On Earth, it would take a little over nine seconds for the ball to fall back to the ground.

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math word problem involving quadratic equation

Word Problems on Quadratic Equations by Factoring

We will learn how to solve Word Problems on quadratic equations by factoring.

1.  The product of two numbers is 12. If their sum added to the sum of their squares is 32, find the numbers.

Let the numbers be x and y.

As their product is 12, we get xy = 12 ..................... (i)

According to the question, x + y + x\(^{2}\) + y\(^{2}\) = 32 ..................... (ii)

From (i), y = \(\frac{12}{x}\)

Putting y = \(\frac{12}{x}\) in (ii), we get

x + \(\frac{12}{x}\) + x\(^{2}\) + (\(\frac{12}{x}\))\(^{2}\) = 32

⟹ (x + \(\frac{12}{x}\)) + (x + \(\frac{12}{x}\))\(^{2}\) - 2 x ∙ \(\frac{12}{x}\) = 32

⟹ (x + \(\frac{12}{x}\))\(^{2}\) + (x + \(\frac{12}{x}\)) - 56 = 0

Putting x + \(\frac{12}{x}\) = t,

t\(^{2}\) + t - 56 = 0

⟹ t\(^{2}\) + 8t – 7t – 56 = 0

⟹ t(t + 8) - 7(t + 8) = 0

⟹ (t + 8)(t - 7) = 0

⟹ t + 8 = 0 or, t – 7 = 0

⟹ t = -8 or, t = 7

When t = -8,

x + \(\frac{12}{x}\) = t = -8

⟹ x\(^{2}\) + 8x + 12 = 0

⟹ x\(^{2}\) + 6x + 2x + 12 = 0

⟹ x(x + 6) + 2(x + 6) = 0

⟹ (x + 6)(x + 2) = 0

⟹ x + 6 = 0 or, x + 2 = 0

⟹ x = -6 or, x = -2

x + \(\frac{12}{x}\) = t = 7

⟹ x\(^{2}\) - 7x + 12 = 0

⟹ x\(^{2}\) - 4x - 3x + 12 = 0

⟹ x(x – 4) - 3(x – 4) = 0

⟹ (x - 4)(x - 3) = 0

⟹ x - 4 = 0 or, x - 3 = 0

⟹ x = 4 or 3

Thus, x = -6, -2, 4, 3

Then, the other number y = \(\frac{12}{x}\) = \(\frac{12}{-6}\) , \(\frac{12}{-2}\), \(\frac{12}{4}\), \(\frac{12}{3}\) = -2, -6, 3, 4.

Thus, the two numbers x, y are -6, -2, or -2, -6, or 4, 3 or 3, 4.

Therefore, the required two numbers are -6, -2 or 4, 3.

2. An association has a fund of $195. In addition that, each member of the association contributes the number of dollars equal to the number of members. The total money is divided equally among the members. If each of the members gets $ 28, find the number of members in the association.

Let the number of members be x.

Total contributions from them = $ x\(^{2}\) and the association has a fund of $ 195.

According to the problem,

x\(^{2}\)  + 195 = 28x

⟹ x\(^{2}\)  - 28x + 195 = 0

⟹ x\(^{2}\) - 15x - 13x + 195 = 0

⟹ x(x - 15) - 13(x - 15) = 0

⟹ (x - 15)(x - 13) = 0

Therefore, x = 15 or 13

There are 15 or 13 members in the association.

Note: Two answers are acceptable in this case.

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

Solving Quadratic Equations

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring

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Real World Examples of Quadratic Equations

A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

  • Factoring Quadratics
  • Completing the Square
  • Graphing Quadratic Equations
  • The Quadratic Formula
  • Online Quadratic Equation Solver

Each example follows three general stages:

  • Take the real world description and make some equations
  • Use your common sense to interpret the results

ball throw

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. when does it hit the ground.

Ignoring air resistance, we can work out its height by adding up these three things: (Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t 2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t 2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t 2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1 :

5t 2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give ac , and add to give b " method in Factoring Quadratics :

ac = −15 , and b = −14 .

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t 2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations , and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a :

  • t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

  • h = −5t 2 + 14t + 3 = −5(1.4) 2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Example: New Sports Bike

bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your costs are going to be:

  • $700,000 for manufacturing set-up costs, advertising, etc
  • $110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

  • at $0, you just give away 70,000 bikes
  • at $350, you won't sell any bikes at all
  • at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

Profit = −200P 2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

Solve: −200P 2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = (−460/2) 2 = (−230) 2 = 52900

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230 , and you can expect:

  • Unit Sales = 70,000 − 200 x 230 = 24,000
  • Sales in Dollars = $230 x 24,000 = $5,520,000
  • Costs = 700,000 + $110 x 24,000 = $3,340,000
  • Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm 2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area of steel after cutting out the 11 × 6 middle:

Let us solve this one graphically !

Here is the graph of 4x 2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm 2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. the river has a current of 2 km an hour. what is the boat's speed and how long was the upstream journey.

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

  • Let x = the boat's speed in the water (km/h)
  • Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

  • when going upstream, v = x−2 (its speed is reduced by 2 km/h)
  • when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2) (x+2) :

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x 2 −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x 2 − 30x − 12 = 0

It is a Quadratic Equation!

Let us solve it using the Quadratic Formula :

Where a , b and c are from the Quadratic Equation in "Standard Form": ax 2 + bx + c = 0

Solve 3x 2 - 30x - 12 = 0

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

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Course: Algebra 1   >   Unit 14

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Video transcript

Math Tutor DVD - Online Math Help, Math Homework Help, Math Problems, Math Practice!

  • 1c - Sample Algebra Word Prob
  • Section 11: Algebra Word Problems In . . .

Section 11: Algebra Word Problems Involving Quadratic Equations

This section teaches the student how to solve algebra word problems involving quadratic equations. Every problem in this section will involve a quadratic equation that the student must then solve.

IMAGES

  1. Quadradic Equation Word Problems

    math word problem involving quadratic equation

  2. 3 Ways to Solve Word Problems Requiring Quadratic Equations

    math word problem involving quadratic equation

  3. Quadratic Equation Word Problems Worksheets

    math word problem involving quadratic equation

  4. Using Quadratic Equations to Solve Problems

    math word problem involving quadratic equation

  5. Word Problems Using Quadratic Formula

    math word problem involving quadratic equation

  6. Word Problems Using Quadratic Equations Examples

    math word problem involving quadratic equation

VIDEO

  1. Solving Problems Involving Quadratic Equations

  2. Solving Quadratic Equations: The Basics

  3. Maths-Geometry word problem using Quadratic Equation

  4. Solving Problems Involving Quadratic Equations

  5. O level Math

  6. 10th Maths

COMMENTS

  1. 10 Quadratic Equations Word Problems

    Quadratic equations word problems are math problems in which the equations are not given directly. These problems can be solved by using the given information to obtain a quadratic equation of the form ax^2+bx+c ax2 + bx+ c. We can then use the factoring method, the completing the square method or the quadratic formula to solve the equation.

  2. Word Problems Involving Quadratic Equations

    Word Problems Involving Quadratic Equations Need Help Solving Those Dreaded Word Problems Involving Quadratic Equations? Yes, I know it's tough. You've finally mastered factoring and using the quadratic formula and now you are asked to solve more problems! Except these are even more tough.

  3. Quadratic equations word problem

    Quadratic equations word problem | Algebra (video) | Khan Academy Course: Algebra 1 > Unit 14 Lesson 10: Quadratic standard form Finding the vertex of a parabola in standard form Graphing quadratics: standard form Graph quadratics in standard form Quadratic word problem: ball Quadratic word problems (standard form) Math > Algebra 1 >

  4. 10.7 Quadratic Word Problems: Age and Numbers

    Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you will end up constructing a quadratic equation. To find the solution, you will be required to either factor the quadratic equation or use substitution. Example 10.7.1

  5. Quadratic Equation Word Problems with Solution

    Solution: Let the numbers be x and x + 1 According to the problem, x (x + 1) = 483 => x 2 + x - 483 = 0 => x 2 + 22x - 21x - 483 = 0 => x (x + 22) - 21 (x + 22) = 0 => (x + 22) (x - 21) = 0 => x + 22 = 0 or x - 21 = 0 => x = {-22, 21} Thus, the two consecutive numbers are 21 and 22.

  6. Quadratic word problems (standard form) (practice)

    Quadratic word problems (standard form) Google Classroom. You might need: Calculator. Rui is a professional deep water free diver. His altitude (in meters relative to sea level), x seconds after diving, is modeled by: d ( x) = 1 2 x 2 − 10 x.

  7. Quadradic Equation Word Problems

    Example 1: Word problems involving area. Many quadratic word problems use the area of a rectangle. Here is a graphic to get you started. Here is another example. Singh knows the total area of his garden is 50 m2, but he doesn't know the length or width. Use the diagram below to find the length and width in meters (m).

  8. Word problems: Solving quadratic equations

    Class 10 (Old) > Quadratic equations > Quadratic equations word problems Word problems: Solving quadratic equations Google Classroom Cullen is 10 years younger than Ada. The product of their ages 2 years ago was 39 . Find Ada's present age. years Stuck? Use a hint. Report a problem Do 4 problems

  9. Quadratic Equations Word Problems

    Question 1a: Find two consecutive integers that have a product of 42 Quadratic equations - Solving word problems using factoring of trinomials Question 1b: There are three consecutive integers. The product of the two larger integers is 30. Find the three integers. Show Video Lesson Quadratic Equations - Solving Word problems by Factoring

  10. PDF QUADRATIC WORD PROBLEMS

    HOW TO SOLVE QUADRATIC EQUATIONS: Find the roots: r2 12 r 35 0 Solve for y: y2 11 y 24 0 3. Find the zeroes: x2 5 x 6 0 4. Solve for y: y2 3 y 28 5. Find the roots: x2 x 30 6. Find the zeros: 5w 2 35 7. A plot of land for sale has a width of x ft., and a length that is 8ft less than its width.

  11. SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

    Solving Word Problems Involving Quadratic Equations SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS Problem 1 : If the difference between a number and its reciprocal is ²⁴⁄₅, find the number. Solution : Let x be the required number. Then its reciprocal is ¹⁄ₓ be its reciprocal. 5(x2 - 1) = 24x 5x2 - 5 = 24x 5x2 - 24x - 5 = 0

  12. Word Problems with Quadratic Equations

    At least one of these methods can be used to solve any problem involving a quadratic equation, and which method you choose depends on the kind of problem you are presented with. ... Now substitute the 'a', 'b' and 'c' values into the quadratic formula, and then do the math. The answer is the same as before, 'x' is equal to 10 and -2. Just as ...

  13. Quadratic equations word problems

    A quadratic equation takes the form of are two integers, known as coefficients of Enter a, b and c to find the solutions of the equations. - x - 6 = 0, where a = 1; b=-1; c=-6 I highly recommend the following textbook for both GCSE (9-1) and IGCSE (9-1). The book covers every single topic in depth and offers plenty of questions to practise.

  14. 9.6: Solve Applications of Quadratic Equations

    This is a quadratic equation, rewrite it in standard form. Solve the equation using the Quadratic Formula. Identify the values of \(a, b, c\). Write the Quadratic Formula. Then substitute in the values of \(a,b,c\). Simplify. Figure 9.5.26: Rewrite to show two solutions. Approximate the answer with a calculator. Step 6: Check the answer. The ...

  15. Word Problems Involving Quadratic Equations Flashcards

    The product of two consecutive odd integers is 2,115. Find the integers. Which of the following quadratic equations could be used to solve the word problem? x² + 2x - 2,115 = 0. The square of one number is 23 less than the square of a second number. If the second number is 1 more than the first, what are the two numbers?

  16. 6.7: Applications Involving Quadratic Equations

    Figure 6.7.4. Use the formula A = 1 2bh and the fact that the area is 7 square inches to set up an algebraic equation. A = 1 2b ⋅ h 7 = 1 2b(2b − 3) To avoid fractional coefficients, multiply both sides by 2 and then rewrite the quadratic equation in standard form. Factor and then set each factor equal to zero.

  17. How to solve quadratic projectile-motion problems

    The projectile-motion equation is s(t) = −½ gx2 + v0x + h0, where g is the constant of gravity, v0 is the initial velocity (that is, the velocity at time t = 0 ), and h0 is the initial height of the object (that is, the height at of the object at t = 0, the time of release).

  18. Word Problems on Quadratic Equations by Factoring

    We will learn how to solve Word Problems on quadratic equations by factoring. 1. The product of two numbers is 12. If their sum added to the sum of their squares is 32, find the numbers.

  19. Real World Examples of Quadratic Equations

    Solve: −200P 2 + 92,000P − 8,400,000 = 0. Step 1 Divide all terms by -200. P 2 - 460P + 42000 = 0. Step 2 Move the number term to the right side of the equation: P 2 - 460P = -42000. Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

  20. Quadratic equations word problems (advanced)

    Class 10 (Old) > Quadratic equations > Quadratic equations word problems Quadratic equations word problems (advanced) Google Classroom You might need: Calculator A two-digit number is such that the product of its digits is 14 . When 45 is added to the number, the digits interchange their places. Let the units digit of the initial number be y .

  21. Word Problems Calculator

    Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems. Show more Why users love our Word Problems Calculator Middle School Math Solutions - Inequalities Calculator

  22. Quadratic equations word problem: triangle dimensions

    Any quadratic of the form ax^2 + bx + c can be solved using the formula ( -b +/- sqrt (D) )/2a with D= b^2-4*a*c. However, if D is less than zero it cannot be solved regularly. This requires the introduction if the imaginary number i = sqrt (-1). I think this allows us to factor all quadratics. So short answer: yes.

  23. Section 11: Algebra Word Problems Involving Quadratic Equations

    This section teaches the student how to solve algebra word problems involving quadratic equations. Every problem in this section will involve a quadratic equation that the student must then solve. Math Tutor DVD. Home ... and examples on the math tutor DVD's are very clearly explained, and Jason's style of teaching definitely makes the viewer ...

  24. PDF Praxis Mathematics (5165) Study Companion

    b. Solves quadratic equations with real coefficients that have complex solutions c. Uses the method of completing the square to transform any quadratic equation in x into the equivalent form () x p q −= 2 d. Solves equations using a variety of methods (e.g., graphing, factoring, using the quadratic formula) e. Uses different methods (e.g.,

  25. Students' Polya problem solving skills on system of linear equations

    This study seeks students' abilities to solve Polya problems involving systems of linear equations with two variables based on their mathematical prowess. Three junior high school students with high, average, and low math aptitudes were chosen as participants. Research data were collected using a written test on the system of linear equations with two variables problem in the form of word ...