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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].

how to solve quadratic equations word problems

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.

how to solve quadratic equations word problems

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.

how to solve quadratic equations word problems

[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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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

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How to Solve Word Problems Requiring Quadratic Equations

Last Updated: December 27, 2020

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 14,258 times.

Some word problems require quadratic equations in order to be solved. In this article, you will learn how to solve those types of problems. Once you get the hang of it, it will be very easy.

Quadratic Equations

Step 1 Know what kind of problem you're tackling.

  • For the real life scenarios, factoring method is better.
  • In geometric problems, it is good to use the quadratic formula.

Real Life Scenario

Step 1 Ask to yourself,

  • In this problem, it asks for Kenny's birthday.

Step 2 Decide your variables.

  • Since negative month does not exist, 3 is the only one that makes sense.
  • Because the problem asks for both the month and the date, the answer would be March 18th. (Use the value for the other variable that you found in step 3.)

Geometric Problems

Step 1 Identify if it's a geometric problem.

  • In the problem above, it asks you only for the height of the triangle.

Step 3 Decide your variables.

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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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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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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 Quadratic Equations

Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

Step I: Denote the unknown quantities by x, y etc.

Step II: use the conditions of the problem to establish in unknown quantities.

Step III: Use the equations to establish one quadratic equation in one unknown.

Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

Now we will learn how to frame the equations from word problem:

1.  The product of two consecutive integers is 132. Frame an equation for the statement. What is the degree of the equation?

Method I: Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

Therefore, x(x + 1) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

Method II: Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

From (i), y = 1 + x.

Putting y = 1 + x in (ii),

x(1 + x) = 132

Solving the quadratic equation, we get the value of x. Then the value of y can be determined by substituting the value of x in y = 1 + x.

2. The length of a rectangle is greater than its breadth by 3m. If its area be 10 sq. m, find the perimeter.

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

⟹ x\(^{2}\) + 3x - 10 = 0

⟹ (x + 5)(x - 2) = 0

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m

and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = -5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

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 

Word Problems on Quadratic Equations by Factoring

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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General Quadratic Word Problems

Projectiles General Word Problems Max/Min Problems

Most quadratic word problems should seem very familiar, as they are built from the linear problems that you've done in the past. You will need to use keywords to interpret the English and, from that, create the quadratic model. Then solve the model for the solutions (that is, the x -intercepts).

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When interpreted — within the context of the exercise — one or both of the solutions to the quadratic equation will provide at least part of the answer they've asked of you.

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  • A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?

The height is defined in terms of the width, so I'll pick a variable for "width", and then create an expression for the height.

Let " w " stand for the width of the picture. The height h is given as being 4/3 of the width, so:

h = (4/3) w

Then the area, being the product of the width and the height, is given by:

= [(4/3) w ][ w ]

= (4/3) w 2 = 192

I need to solve this "area" equation for the value of the width, and then back-solve to find the value of the height.

(4/3) w 2 = 192 w 2 = 144 w = ± 12

Since I can't have a negative width, I can ignore the " w  = −12 " solution. Then the width must be the other solution, 12 , and the height is then:

h = (4/3)(12) = 16

I can do a quick check of my answer, since this is a "solving" exercise, by plugging my values back into the original exercise. Since 12 × 16 does indeed equal 192 , my answer checks. So my hand-in answer is:

The enlargement will be 12  inches by 16  inches.

  • The product of two consecutive negative integers is 1122 . What are the numbers?

Remember that consecutive integers are one unit apart, so my numbers are n and n  + 1 . Multiplying to get the product, I get:

n ( n + 1) = 1122 n 2 + n = 1122 n 2 + n − 1122 = 0 ( n + 34)( n − 33) = 0

The solutions are n  = −34 and n  = 33 . I need a negative value (because the exercise statement specified that the numbers are negative), so I'll ignore the n  = 33 & and take n  = −34 . Then the other number is:

n + 1 = (−34) + 1 = −33

Then my hand-in answer is:

The two numbers are −33 and −34 .

Note that the second value could have been gotten by changing the sign on the extraneous solution. Warning: Many students get in the very bad habit of arbitrarily changing signs to get the answers they need, but this does not always work, and will very likely get them in trouble later on. (And it'll cause trouble right now, if the grader is paying attention.)

Take the extra half a second to find the right answer the right way.

  • A garden measuring 12 meters by 16 meters is to have a pedestrian pathway installed all around it, increasing the total area to 285 square meters. What will be the width of the pathway?

The first thing I need to do is draw a picture. Since I don't know how wide the path will be, I'll label the width as " x ".

Looking at my picture, I see that the total width of the finished garden-plus-path area will be:

x + 12 + x = 12 + 2 x

...and the total length will be:

x + 16 + x = 16 + 2 x

Then the new area, being the product of the new width and height, is given by:

(12 + 2 x )(16 + 2 x ) = 285 192 + 56 x + 4 x 2 = 285 4 x 2 + 56 x − 93 = 0

This quadratic is messy enough that I won't bother with trying to use factoring to solve; I'll just go straight to the Quadratic Formula :

Obviously the negative value won't work in this context, so I'll ignore it. Checking the original exercise to verify what I'm being asked to find, I notice that I need to have units on my answer:

The width of the pathway will be 1.5 meters.

Algebra Tutors

  • You have to make a square-bottomed, unlidded box with a height of three inches and a volume of approximately 42 cubic inches. You will be taking a piece of cardboard, cutting three-inch squares from each corner, scoring between the corners, and folding up the edges. What should be the dimensions of the cardboard, to the nearest quarter inch?

When dealing with geometric sorts of word problems, it is usually helpful to draw a picture. Since I'll be cutting equal-sized squares out of all of the corners, and since the box will have a square bottom, I know I'll be starting with a square piece of cardboard.

I don't know how big the cardboard will be yet, so I'll label the sides as having length " w ".

Since I know I'll be cutting out three-by-three squares to get sides that are three inches high, I can mark that on my drawing.

The dashed (rather than solid) lines show where I'll be scoring the cardboard and folding up the sides.

Since I'll be losing three inches on either end of the cardboard when I fold up the sides, the final width of the bottom will be the original " w " inches, less three on the one side and another three on the other side. That is, the width of the bottom will be:

w − 3 − 3 = w − 6

The volume will be the product of the width, the length (which is the same as the width), and the depth (which was created by cutting out the corners and folding up the sides).

Then the volume of the box, working from the drawing, gives me the following equation:

( w − 6)( w − 6)(3) = 42 ( w − 6)( w − 6) = 14 ( w − 6) 2 = 14

This is the quadratic I need to solve. I can take the square root of either side, and then add the to the right-hand side:

...or I can multiply out the square and apply the Quadratic Formula :

w 2 − 12 w + 36 = 14

w 2 − 12 w + 22 = 0

Either way, I get two solutions which, when expressed in practical decimal terms, tell me that the width of the original cardboard is either about 2.26 inches or else about 9.74 inches.

How do I know which solution value for the width is right? By checking each value in the original word problem.

If the cardboard is only 2.26 inches wide, then how on earth would I be able to fold up three-inch-deep sides? But if the cardboard is 9.74 inches, then I can fold up three inches of cardboard on either side, and still be left with 3.74 inches in the middle. Checking:

(3.74)(3.74)(3) = 41.9628

This isn't exactly 42 , but, taking round-off error into account, it's close enough that I can trust that I have the correct value. So my hand-in answer, complete with units, is:

The cardboard should measure 9.75  inches on a side.

In this last exercise above, you should notice that each solution method (factoring and the Quadratic Formula) gave the same final answer for the cardboard's width. But the Quadratic Formula took longer and provided me with more opportunities to make mistakes.

Moral? Don't get stuck in the rut of always using the Quadratic Formula.

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how to solve quadratic equations word problems

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

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "R T " is:

1 R T   =   1 R 1 + 1 R 2

In this case, we have R T = 2 and R 2 = R 1 + 3

1 2   =   1 R 1 + 1 R 1 +3

To get rid of the fractions we can multiply all terms by 2R 1 (R 1 + 3) and then simplify:

Yes! A Quadratic Equation!

Let us solve it using our Quadratic Equation Solver .

  • Enter 1, −1 and −6
  • And you should get the answers −2 and 3

R 1 cannot be negative, so R 1 = 3 Ohms is the answer.

The two resistors are 3 ohms and 6 ohms.

Quadratic Equations are useful in many other areas:

parabolic dish

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

And many questions involving time, distance and speed need quadratic equations.

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Course: Algebra 1   >   Unit 14

  • Zero product property
  • Graphing quadratics in factored form
  • Graph quadratics in factored form

Quadratic word problems (factored form)

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Video transcript

IMAGES

  1. 3 Ways to Solve Word Problems Requiring Quadratic Equations

    how to solve quadratic equations word problems

  2. Quadratic Equation Word Problem Example Height of a ball

    how to solve quadratic equations word problems

  3. Quadradic Equation Word Problems

    how to solve quadratic equations word problems

  4. Word Problems

    how to solve quadratic equations word problems

  5. Using Quadratic Equations to Solve Problems

    how to solve quadratic equations word problems

  6. Quadratic equation word problems

    how to solve quadratic equations word problems

VIDEO

  1. QUADRATIC EQUATIONS

  2. Quadratic Equations Revision Extra Word Problems for CBSE class 10 @StudyPointPro

  3. Quadratic equations Extra questions word problems Part 4

  4. QUADRATIC EQUATIONS

  5. Word Problems on Quadratic Equations Part 2

  6. Word Problems on Quadratic Equations Part 1

COMMENTS

  1. Quadratic equations word problem

    Quadratic word problems (standard form) Math > Algebra 1 > Quadratic functions & equations > ... I know its a exercise about quadratic equation not kinematic, but i it can confuse. Just look: s = Vavg. ... So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. And if you want to simplify this a little bit ...

  2. 10.7 Quadratic Word Problems: Age and Numbers

    The equation to use for this problem is (x+ 2)2 − (x)2 = 68 ( x + 2) 2 − ( x) 2 = 68. Simplifying this yields: (x + 2)2 − (x)2 = 68 x2 + 4x + 4 − x2 = 68 4x + 4 = 68 − 4 −4 4x 4 = 64 4 x = 16 ( x + 2) 2 − ( x) 2 = 68 x 2 + 4 x + 4 − x 2 = 68 4 x + 4 = 68 − 4 − 4 4 x 4 = 64 4 x = 16 This means that the two integers are 16 and 18. Example 10.7.3

  3. Word Problems Involving Quadratic Equations

    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.

  4. 3 Ways to Solve Word Problems Requiring Quadratic Equations

    Method 1 Quadratic Equations Know what kind of problem you're tackling. Quadratic equations can be in many forms. In this article, we will use where a≠ 0. You can solve a quadratic equations using the quadratic formula or factoring. For the real life scenarios, factoring method is better.

  5. 10 Quadratic Equations Word Problems

    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. Here, we will look at 10 quadratic equations word problems with answers.

  6. 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.

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

    Algebra 1 > Quadratic functions & equations > Quadratic standard form 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

  8. 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.

  9. APPLICATIONS FOR QUADRATIC EQUATIONS

    Learn how to solve the most common word problems for applications of quadratic equations with this step-by-step tutorial! We'll go through how to solve for w...

  10. QUADRATIC EQUATIONS: word problem

    How to solve quadratic equation word problems - example. For more in-depth math help check out my catalog of courses. Every course includes over 275 videos o...

  11. Word Problems with Quadratic Equations

    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.

  12. Quadratic Equation Word Problems

    This video covers how to solve quadratic equation word problems. In these examples, I show how to translate the verbal statements into mathematical expressio...

  13. IXL

    Test prep Awards Improve your math knowledge with free questions in "Solve quadratic equations: word problems" and thousands of other math skills.

  14. 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

  15. Quadratic functions & equations

    Familiar Attempted Not started Quiz Unit test About this unit We've seen linear and exponential functions, and now we're ready for quadratic functions. We'll explore how these functions and the parabolas they produce can be used to solve real-world problems. Intro to parabolas Learn Parabolas intro Interpreting a parabola in context

  16. Solving Quadratic Equations

    Methods of solving quadratic equations are discussed here in the following steps. Step II: use the conditions of the problem to establish in unknown quantities. Step III: Use the equations to establish one quadratic equation in one unknown. Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

  17. Word Problems on Quadratic Equation: Various Methods

    We can solve any word problems on a quadratic equation using various methods. Let us know about these. Solving Quadratic Equation Sums Using Factorisation Method If we can factorize a x 2 + b x + c, a ≠ 0, into a product of two linear factors, then the roots of the quadratic equation a x 2 + b x + c = 0 can be found by equating each factor to zero.

  18. Quadratic equations, paths, gardens, and boxes

    Most quadratic word problems should seem very familiar, as they are built from the linear problems that you've done in the past. You will need to use keywords to interpret the English and, from that, create the quadratic model. Then solve the model for the solutions (that is, the x-intercepts).

  19. Solving Quadratic Equations by the Quadratic Formula

    Example 5: Solve the quadratic equation below using the Quadratic Formula. First, we need to rewrite the given quadratic equation in Standard Form, [latex]a {x^2} + bx + c = 0 [/latex]. Eliminate the [latex] {x^2} [/latex] term on the right side. Eliminate the [latex]x [/latex] term on the right side. Eliminate the constant on the right side.

  20. 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:

  21. 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

  22. Using the Quadratic Formula to Solve Equations with Literal

    The quadratic formula is not only used for solving quadratic equations but also can be used with literal coefficients, or letters instead of...

  23. 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?

  24. Quadratic word problems (factored form) (video)

    If you start from - 4(x+2)(x-18) and expand it to f(x) = - 4(x^2 - 16x - 36) or f(x) = -4x^2 + 64x + 144, the -4 has to do with the force of gravity trying to pull the rocket back to the ground, the 64 has to do with the initial velocity that the rocket was launched at, and the 144 has to do with the initial height of the rocket.

  25. Solved solving a word problem using a quadratic equation

    Question: solving a word problem using a quadratic equation with rational roots The length of a rectangle is 3 ft less than twice the width, and the area of the rectangle is 14 ft square root 2. Find the dimensions of the rectangle.