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Course: Algebra 1   >   Unit 14

• 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)

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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.

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• EXPLORE Random Article

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

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

Real Life Scenario

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

• 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

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

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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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Quadratic Word Problems

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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:

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

What are Quadratic Functions?

Graphing Quadratic Functions

FOILing and Explanation for FOIL

Solving Quadratic Equations by Taking Square Roots

Solving Quadratic Equations by Factoring

Factoring with Grouping

Solving Quadratic Equations Using the Quadratic Formula

Solving Quadratic Equations by Completing the Square

Finding the Value that Completes the Square

Using and Understanding the Discriminant

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HOW TO SOLVE WORD PROBLEMS IN QUADRATIC EQUATIONS

We can follow the steps given below to solve word problems using quadratic equations.

Understanding the question is more important than any other thing. That is, always it is very important to understand the information given in the question rather than solving.

If it is possible, we have to split the given information. Because, when we split the given information in to parts, we can understand them easily.

Once we understand the given information clearly, solving the word problem in quadratic equation would not be a challenging work. 

When we try to solve the word problems in quadratic equations, we have to introduce "x" or some other alphabet for unknown value (=answer for our question) and form a quadratic equation with this "x". Finally we have to get value for the alphabet which was introduced for the unknown value.

If it is required, we have to draw picture for the given information. Drawing picture for the given information will give us a clear understanding about the question.

Using the alphabet introduced for unknown value, we have to translate the English statement (information) given in the question as quadratic equation equation.

In translation, we have to translate  the following English words as the corresponding mathematical symbols.

of -----> x (multiplication)

am, is, are, was, were, will be, would be --------> = (equal)

Once we have translated the English Statement (information) given in the question as quadratic equation correctly, 90% of the work will be over. The remaining 10% is just getting the answer. That is solving for the unknown. 

These are the steps most commonly involved in solving word problems in quadratic equations.

Let us see how the above explained steps work in solving word problems using quadratic equations. 

A piece of iron rod cost $ 60. If the rod was 2 meter shorter and each meter costs $ 1 more and the total cost  would remain unchanged. What is the length of the rod?  

Let us understand the given information. There are three information given in the question. 

1.  A piece of iron rod costs $ 60.

2.  If the rod was 2 meter shorter and each meter costs $ 1 more

3.  Total cost  would remain unchanged.

Target of the question : What is the length of the rod?

Step 3 :

Let "x" be the length of the rod.

Clearly, we have to find the value of "x" 

If the rod is 2 meter shorter, length of the rod is

=  (x-2)

From the third information, we have the following statements.

Total cost of rod having length x meters is $ 60.

Total cost of rod having length (x-2) meters is $ 60.

Cost of 1 meter of rod having length x meters is

=  60 / x  -----(1)

Cost of 1 meter of rod having length (x-2) meters is

=  60 / (x - 2) -----(2)

From the second information, we can consider the following example. 

That is, if the cost of 1 meter of rod x is $10, then the cost of 1 meter of rod (x-2) will be $11.  

$10 &  $11 can be balanced as shown below.

10 + 1  =  11

(This is just for en example)

If we apply the same logic for (1) & (2), we get 

(60 / x) + 1  =  60 / (x - 2)

(60 + x) / x  =  60 / (x - 2)

(x + 60)(x - 2)  =  60x

x 2  + 58x - 120  =  60x

x 2  - 2x - 120  =  0

(x - 12)(x + 10)  =  0 

x - 12  =  0     or     x + 10  =  0

x  =  12     or     x  =  -10

Because length can never be a negative value, we can ignore x  =  -10. 

x  =  12

So, the length of the rod is 12 meter. 

Would you like to practice more word problems in quadratic equations ? 

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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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