assignment on quadratic equation

Quadratic Formula Exercises

Quadratic formula practice problems with answers.

Below are ten (10) practice problems regarding the quadratic formula. The more you use the formula to solve quadratic equations, the more you become expert at it!

Use the illustration below as a guide. Notice that in order to apply the quadratic formula, we must transform the quadratic equation into the standard form, that is, [latex]a{x^2} + bx + c = 0[/latex] where [latex]a \ne 0[/latex].

The problems below have varying levels of difficulty. I encourage you to try them all. Believe me, they are actually easy! Good luck.

Problem 1: Solve the quadratic equation using the quadratic formula.

[latex]{x^2}\, – \,8x + 12 = 0[/latex]

Therefore, the answers are [latex]{x_1} = 6[/latex] and [latex]{x_2} = 2[/latex].

Problem 2: Solve the quadratic equation using the quadratic formula.

[latex]2{x^2}\, -\, x = 1[/latex]

Rewrite the quadratic equation in the standard form.

[latex]2{x^2} – x – 1 = 0[/latex]

Therefore, the answers are [latex]{x_1} = 1[/latex] and [latex]{x_2} = \large{{ – 1} \over 2}[/latex].

Problem 3: Solve the quadratic equation using the quadratic formula.

[latex]4{x^2} + 9 = – 12x[/latex]

[latex]4{x^2} + 12x + 9 = 0[/latex]

Therefore, the solution is [latex]x = \large{{ – 3} \over 2}[/latex].

Problem 4: Solve the quadratic equation using the quadratic formula.

[latex]5{x^2} = 7x + 6[/latex]

Convert the quadratic equation into the standard form.

[latex]5{x^2} – 7x – 6 = 0[/latex]

Therefore, the answers are [latex]{x_1} = 2[/latex] and [latex]{x_2} = \large{{ – 3} \over 5}[/latex].

Problem 5: Solve the quadratic equation using the quadratic formula.

[latex]{x^2} -\,{ \large{1 \over 2}}x\, – \,{\large{3 \over {16}}} = 0[/latex]

Multiply the entire equation by the LCM of the denominators which is [latex]16[/latex]. This will get rid of the denominators thereby giving us integer values for [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex].

[latex]16{x^2} – 8x – 3 = 0[/latex]

Therefore, the answers are [latex]x_1=\large{3 \over 4}[/latex] and [latex]x_2=\large{{ – 1} \over 4}[/latex].

Problem 6: Solve the quadratic equation using the quadratic formula.

[latex]{x^2} + 3x + 9 = 5x – 8[/latex]

Convert into standard form as [latex]{x^2} – 2x + 17 = 0[/latex].

Therefore, the answers are [latex]x_1=1 + 4i[/latex] and [latex]x_2=1 – 4i[/latex].

Problem 7: Solve the quadratic equation using the quadratic formula.

[latex]{\left( {x – 2} \right)^2} = 4x[/latex]

Rewrite in standard form as [latex]{x^2} – 8x + 4 = 0[/latex].

Hence, the answers are [latex]{x_1} = 4 + 2\sqrt 3 [/latex] and [latex]{x_2} = 4 – 2\sqrt 3 [/latex].

Problem 8: Solve the quadratic equation using the quadratic formula.

[latex]{\Large{{{x^2}} \over 4} – {x \over 2} }= 1[/latex]

To convert the quadratic equation into the standard form, simply multiply the entire equation by [latex]4[/latex] then subtract both sides by [latex]4[/latex].

[latex]{x^2} – 2x – 4 = 0[/latex]

Thus, the answers are [latex]{x_1} = 1 + \sqrt 5 [/latex] and [latex]{x_2} = 1 – \sqrt 5 [/latex].

Problem 9: Solve the quadratic equation using the quadratic formula.

[latex]{\left( {2x – 1} \right)^2} = \Large{x \over 3}[/latex]

If we carefully transform the given quadratic equation into the standard form, we get [latex]12{x^2} – 13x + 3 = 0[/latex].

Therefore, the answers are [latex]x_1={\Large{3 \over 4}}[/latex] and [latex]x_2={\Large{1 \over 3}}[/latex].

Problem 10: Solve the quadratic equation using the quadratic formula.

[latex]\left( {2x – 1} \right)\left( {x + 4} \right) = – {x^2} + 3x[/latex]

If we simplify the quadratic equation to convert it to the standard form, we should arrive at [latex]3{x^2} + 4x – 4 = 0[/latex].

Hence, the answers are [latex]x_1={\Large{2 \over 3}}[/latex] and [latex]x_2=-2[/latex].

You may also be interested in these related math lessons or tutorials:

The Quadratic Formula

Solving Quadratic Equations using the Quadratic Formula

9.3 Solve Quadratic Equations Using the Quadratic Formula

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic equations using the Quadratic Formula
• Use the discriminant to predict the number and type of solutions of a quadratic equation
• Identify the most appropriate method to use to solve a quadratic equation

Be Prepared 9.7

Before you get started, take this readiness quiz.

Evaluate b 2 − 4 a b b 2 − 4 a b when a = 3 a = 3 and b = −2 . b = −2 . If you missed this problem, review Example 1.21 .

Be Prepared 9.8

Simplify: 108 . 108 . If you missed this problem, review Example 8.13 .

Be Prepared 9.9

Simplify: 50 . 50 . If you missed this problem, review Example 8.76 .

Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for x.

We start with the standard form of a quadratic equation and solve it for x by completing the square.

• Quadratic Formula

The solutions to a quadratic equation of the form ax 2 + bx + c = 0, where a ≠ 0 a ≠ 0 are given by the formula:

To use the Quadratic Formula , we substitute the values of a , b , and c from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.

Notice the formula is an equation. Make sure you use both sides of the equation.

Example 9.21

How to solve a quadratic equation using the quadratic formula.

Solve by using the Quadratic Formula: 2 x 2 + 9 x − 5 = 0 . 2 x 2 + 9 x − 5 = 0 .

Try It 9.41

Solve by using the Quadratic Formula: 3 y 2 − 5 y + 2 = 0 3 y 2 − 5 y + 2 = 0 .

Try It 9.42

Solve by using the Quadratic Formula: 4 z 2 + 2 z − 6 = 0 4 z 2 + 2 z − 6 = 0 .

Solve a quadratic equation using the quadratic formula.

• Step 1. Write the quadratic equation in standard form, ax 2 + bx + c = 0. Identify the values of a , b , and c .
• Step 2. Write the Quadratic Formula. Then substitute in the values of a , b , and c .
• Step 3. Simplify.
• Step 4. Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “ x =”.

Example 9.22

Solve by using the Quadratic Formula: x 2 − 6 x = −5 . x 2 − 6 x = −5 .

Try It 9.43

Solve by using the Quadratic Formula: a 2 − 2 a = 15 a 2 − 2 a = 15 .

Try It 9.44

Solve by using the Quadratic Formula: b 2 + 24 = −10 b b 2 + 24 = −10 b .

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula . If we get a radical as a solution, the final answer must have the radical in its simplified form.

Example 9.23

Solve by using the Quadratic Formula: 2 x 2 + 10 x + 11 = 0 . 2 x 2 + 10 x + 11 = 0 .

Try It 9.45

Solve by using the Quadratic Formula: 3 m 2 + 12 m + 7 = 0 3 m 2 + 12 m + 7 = 0 .

Try It 9.46

Solve by using the Quadratic Formula: 5 n 2 + 4 n − 4 = 0 5 n 2 + 4 n − 4 = 0 .

When we substitute a , b , and c into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

Example 9.24

Solve by using the Quadratic Formula: 3 p 2 + 2 p + 9 = 0 . 3 p 2 + 2 p + 9 = 0 .

Try It 9.47

Solve by using the Quadratic Formula: 4 a 2 − 2 a + 8 = 0 4 a 2 − 2 a + 8 = 0 .

Try It 9.48

Solve by using the Quadratic Formula: 5 b 2 + 2 b + 4 = 0 5 b 2 + 2 b + 4 = 0 .

Remember, to use the Quadratic Formula, the equation must be written in standard form, ax 2 + bx + c = 0. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example 9.25

Solve by using the Quadratic Formula: x ( x + 6 ) + 4 = 0 . x ( x + 6 ) + 4 = 0 .

Our first step is to get the equation in standard form.

Try It 9.49

Solve by using the Quadratic Formula: x ( x + 2 ) − 5 = 0 . x ( x + 2 ) − 5 = 0 .

Try It 9.50

Solve by using the Quadratic Formula: 3 y ( y − 2 ) − 3 = 0 . 3 y ( y − 2 ) − 3 = 0 .

When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.

Example 9.26

Solve by using the Quadratic Formula: 1 2 u 2 + 2 3 u = 1 3 . 1 2 u 2 + 2 3 u = 1 3 .

Our first step is to clear the fractions.

Try It 9.51

Solve by using the Quadratic Formula: 1 4 c 2 − 1 3 c = 1 12 1 4 c 2 − 1 3 c = 1 12 .

Try It 9.52

Solve by using the Quadratic Formula: 1 9 d 2 − 1 2 d = − 1 3 1 9 d 2 − 1 2 d = − 1 3 .

Think about the equation ( x − 3) 2 = 0. We know from the Zero Product Property that this equation has only one solution, x = 3.

We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to 0 gives just one solution. Notice that once the radicand is simplified it becomes 0 , which leads to only one solution.

Example 9.27

Solve by using the Quadratic Formula: 4 x 2 − 20 x = −25 . 4 x 2 − 20 x = −25 .

Did you recognize that 4 x 2 − 20 x + 25 is a perfect square trinomial. It is equivalent to (2 x − 5) 2 ? If you solve 4 x 2 − 20 x + 25 = 0 by factoring and then using the Square Root Property, do you get the same result?

Try It 9.53

Solve by using the Quadratic Formula: r 2 + 10 r + 25 = 0 . r 2 + 10 r + 25 = 0 .

Try It 9.54

Solve by using the Quadratic Formula: 25 t 2 − 40 t = −16 . 25 t 2 − 40 t = −16 .

Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant .

Discriminant

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.

Using the Discriminant, b 2 − 4 ac , to Determine the Number and Type of Solutions of a Quadratic Equation

For a quadratic equation of the form ax 2 + bx + c = 0, a ≠ 0 , a ≠ 0 ,

• If b 2 − 4 ac > 0, the equation has 2 real solutions.
• if b 2 − 4 ac = 0, the equation has 1 real solution.
• if b 2 − 4 ac < 0, the equation has 2 complex solutions.

Example 9.28

Determine the number of solutions to each quadratic equation.

ⓐ 3 x 2 + 7 x − 9 = 0 3 x 2 + 7 x − 9 = 0 ⓑ 5 n 2 + n + 4 = 0 5 n 2 + n + 4 = 0 ⓒ 9 y 2 − 6 y + 1 = 0 . 9 y 2 − 6 y + 1 = 0 .

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

Since the discriminant is positive, there are 2 real solutions to the equation.

Since the discriminant is negative, there are 2 complex solutions to the equation.

Since the discriminant is 0, there is 1 real solution to the equation.

Try It 9.55

Determine the numberand type of solutions to each quadratic equation.

ⓐ 8 m 2 − 3 m + 6 = 0 8 m 2 − 3 m + 6 = 0 ⓑ 5 z 2 + 6 z − 2 = 0 5 z 2 + 6 z − 2 = 0 ⓒ 9 w 2 + 24 w + 16 = 0 . 9 w 2 + 24 w + 16 = 0 .

Try It 9.56

Determine the number and type of solutions to each quadratic equation.

ⓐ b 2 + 7 b − 13 = 0 b 2 + 7 b − 13 = 0 ⓑ 5 a 2 − 6 a + 10 = 0 5 a 2 − 6 a + 10 = 0 ⓒ 4 r 2 − 20 r + 25 = 0 . 4 r 2 − 20 r + 25 = 0 .

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We summarize the four methods that we have used to solve quadratic equations below.

Methods for Solving Quadratic Equations

• Square Root Property
• Completing the Square

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is a x 2 = k a x 2 = k or a ( x − h ) 2 = k a ( x − h ) 2 = k we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.

Identify the most appropriate method to solve a quadratic equation.

• Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
• Step 2. Try the Square Root Property next. If the equation fits the form a x 2 = k a x 2 = k or a ( x − h ) 2 = k , a ( x − h ) 2 = k , it can easily be solved by using the Square Root Property.
• Step 3. Use the Quadratic Formula . Any other quadratic equation is best solved by using the Quadratic Formula.

The next example uses this strategy to decide how to solve each quadratic equation.

Example 9.29

Identify the most appropriate method to use to solve each quadratic equation.

ⓐ 5 z 2 = 17 5 z 2 = 17 ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0 ⓒ 8 u 2 + 6 u = 11 . 8 u 2 + 6 u = 11 .

ⓐ 5 z 2 = 17 5 z 2 = 17

Since the equation is in the a x 2 = k , a x 2 = k , the most appropriate method is to use the Square Root Property.

ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.

Try It 9.57

ⓐ x 2 + 6 x + 8 = 0 x 2 + 6 x + 8 = 0 ⓑ ( n − 3 ) 2 = 16 ( n − 3 ) 2 = 16 ⓒ 5 p 2 − 6 p = 9 . 5 p 2 − 6 p = 9 .

Try It 9.58

ⓐ 8 a 2 + 3 a − 9 = 0 8 a 2 + 3 a − 9 = 0 ⓑ 4 b 2 + 4 b + 1 = 0 4 b 2 + 4 b + 1 = 0 ⓒ 5 c 2 = 125 . 5 c 2 = 125 .

Access these online resources for additional instruction and practice with using the Quadratic Formula.

• Using the Quadratic Formula
• Solve a Quadratic Equation Using the Quadratic Formula with Complex Solutions
• Discriminant in Quadratic Formula

Section 9.3 Exercises

Practice makes perfect.

In the following exercises, solve by using the Quadratic Formula.

4 m 2 + m − 3 = 0 4 m 2 + m − 3 = 0

4 n 2 − 9 n + 5 = 0 4 n 2 − 9 n + 5 = 0

2 p 2 − 7 p + 3 = 0 2 p 2 − 7 p + 3 = 0

3 q 2 + 8 q − 3 = 0 3 q 2 + 8 q − 3 = 0

p 2 + 7 p + 12 = 0 p 2 + 7 p + 12 = 0

q 2 + 3 q − 18 = 0 q 2 + 3 q − 18 = 0

r 2 − 8 r = 33 r 2 − 8 r = 33

t 2 + 13 t = −40 t 2 + 13 t = −40

3 u 2 + 7 u − 2 = 0 3 u 2 + 7 u − 2 = 0

2 p 2 + 8 p + 5 = 0 2 p 2 + 8 p + 5 = 0

2 a 2 − 6 a + 3 = 0 2 a 2 − 6 a + 3 = 0

5 b 2 + 2 b − 4 = 0 5 b 2 + 2 b − 4 = 0

x 2 + 8 x − 4 = 0 x 2 + 8 x − 4 = 0

y 2 + 4 y − 4 = 0 y 2 + 4 y − 4 = 0

3 y 2 + 5 y − 2 = 0 3 y 2 + 5 y − 2 = 0

6 x 2 + 2 x − 20 = 0 6 x 2 + 2 x − 20 = 0

2 x 2 + 3 x + 3 = 0 2 x 2 + 3 x + 3 = 0

2 x 2 − x + 1 = 0 2 x 2 − x + 1 = 0

8 x 2 − 6 x + 2 = 0 8 x 2 − 6 x + 2 = 0

8 x 2 − 4 x + 1 = 0 8 x 2 − 4 x + 1 = 0

( v + 1 ) ( v − 5 ) − 4 = 0 ( v + 1 ) ( v − 5 ) − 4 = 0

( x + 1 ) ( x − 3 ) = 2 ( x + 1 ) ( x − 3 ) = 2

( y + 4 ) ( y − 7 ) = 18 ( y + 4 ) ( y − 7 ) = 18

( x + 2 ) ( x + 6 ) = 21 ( x + 2 ) ( x + 6 ) = 21

1 3 m 2 + 1 12 m = 1 4 1 3 m 2 + 1 12 m = 1 4

1 3 n 2 + n = − 1 2 1 3 n 2 + n = − 1 2

3 4 b 2 + 1 2 b = 3 8 3 4 b 2 + 1 2 b = 3 8

1 9 c 2 + 2 3 c = 3 1 9 c 2 + 2 3 c = 3

16 c 2 + 24 c + 9 = 0 16 c 2 + 24 c + 9 = 0

25 d 2 − 60 d + 36 = 0 25 d 2 − 60 d + 36 = 0

25 q 2 + 30 q + 9 = 0 25 q 2 + 30 q + 9 = 0

16 y 2 + 8 y + 1 = 0 16 y 2 + 8 y + 1 = 0

Use the Discriminant to Predict the Number of Real Solutions of a Quadratic Equation

In the following exercises, determine the number of real solutions for each quadratic equation.

ⓐ 4 x 2 − 5 x + 16 = 0 4 x 2 − 5 x + 16 = 0 ⓑ 36 y 2 + 36 y + 9 = 0 36 y 2 + 36 y + 9 = 0 ⓒ 6 m 2 + 3 m − 5 = 0 6 m 2 + 3 m − 5 = 0

ⓐ 9 v 2 − 15 v + 25 = 0 9 v 2 − 15 v + 25 = 0 ⓑ 100 w 2 + 60 w + 9 = 0 100 w 2 + 60 w + 9 = 0 ⓒ 5 c 2 + 7 c − 10 = 0 5 c 2 + 7 c − 10 = 0

ⓐ r 2 + 12 r + 36 = 0 r 2 + 12 r + 36 = 0 ⓑ 8 t 2 − 11 t + 5 = 0 8 t 2 − 11 t + 5 = 0 ⓒ 3 v 2 − 5 v − 1 = 0 3 v 2 − 5 v − 1 = 0

ⓐ 25 p 2 + 10 p + 1 = 0 25 p 2 + 10 p + 1 = 0 ⓑ 7 q 2 − 3 q − 6 = 0 7 q 2 − 3 q − 6 = 0 ⓒ 7 y 2 + 2 y + 8 = 0 7 y 2 + 2 y + 8 = 0

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

ⓐ x 2 − 5 x − 24 = 0 x 2 − 5 x − 24 = 0 ⓑ ( y + 5 ) 2 = 12 ( y + 5 ) 2 = 12 ⓒ 14 m 2 + 3 m = 11 14 m 2 + 3 m = 11

ⓐ ( 8 v + 3 ) 2 = 81 ( 8 v + 3 ) 2 = 81 ⓑ w 2 − 9 w − 22 = 0 w 2 − 9 w − 22 = 0 ⓒ 4 n 2 − 10 n = 6 4 n 2 − 10 n = 6

ⓐ 6 a 2 + 14 a = 20 6 a 2 + 14 a = 20 ⓑ ( x − 1 4 ) 2 = 5 16 ( x − 1 4 ) 2 = 5 16 ⓒ y 2 − 2 y = 8 y 2 − 2 y = 8

ⓐ 8 b 2 + 15 b = 4 8 b 2 + 15 b = 4 ⓑ 5 9 v 2 − 2 3 v = 1 5 9 v 2 − 2 3 v = 1 ⓒ ( w + 4 3 ) 2 = 2 9 ( w + 4 3 ) 2 = 2 9

Writing Exercises

Solve the equation x 2 + 10 x = 120 x 2 + 10 x = 120

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?

Solve the equation 12 y 2 + 23 y = 24 12 y 2 + 23 y = 24

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Quadratic Equations

An example of a Quadratic Equation :

The function can make nice curves like this one:

The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x 2 ).

It is also called an "Equation of Degree 2" (because of the "2" on the x )

Standard Form

The Standard Form of a Quadratic Equation looks like this:

• a , b and c are known values. a can't be 0
• x is the variable or unknown (we don't know it yet)

Here are some examples:

Have a Play With It

Play with the Quadratic Equation Explorer so you can see:

• the function's graph, and
• the solutions (called "roots").

Hidden Quadratic Equations!

As we saw before, the Standard Form of a Quadratic Equation is

But sometimes a quadratic equation does not look like that!

For example:

How To Solve Them?

The " solutions " to the Quadratic Equation are where it is equal to zero .

They are also called " roots ", or sometimes " zeros "

There are usually 2 solutions (as shown in this graph).

And there are a few different ways to find the solutions:

Just plug in the values of a, b and c, and do the calculations.

We will look at this method in more detail now.

About the Quadratic Formula

First of all what is that plus/minus thing that looks like ± ?

The ± means there are TWO answers:

x = −b + √(b 2 − 4ac) 2a

x = −b − √(b 2 − 4ac) 2a

Here is an example with two answers:

But it does not always work out like that!

• Imagine if the curve "just touches" the x-axis.
• Or imagine the curve is so high it doesn't even cross the x-axis!

This is where the "Discriminant" helps us ...

Discriminant

Do you see b 2 − 4ac in the formula above? It is called the Discriminant , because it can "discriminate" between the possible types of answer:

• when b 2 − 4ac is positive, we get two Real solutions
• when it is zero we get just ONE real solution (both answers are the same)
• when it is negative we get a pair of Complex solutions

Complex solutions? Let's talk about them after we see how to use the formula.

Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x 2 + 6x + 1 = 0

Answer: x = −0.2 or x = −1

Let's check the answers:

Remembering The Formula

A kind reader suggested singing it to "Pop Goes the Weasel":

Try singing it a few times and it will get stuck in your head!

Or you can remember this story:

x = −b ± √(b 2 − 4ac) 2a

"A negative boy was thinking yes or no about going to a party, at the party he talked to a square boy but not to the 4 awesome chicks. It was all over at 2 am. "

Complex Solutions?

When the Discriminant (the value b 2 − 4ac ) is negative we get a pair of Complex solutions ... what does that mean?

It means our answer will include Imaginary Numbers . Wow!

Example: Solve 5x 2 + 2x + 1 = 0

Answer: x = −0.2 ± 0.4 i

The graph does not cross the x-axis. That is why we ended up with complex numbers.

In a way it is easier: we don't need more calculation, we leave it as −0.2 ± 0.4 i .

Example: Solve x 2 − 4x + 6.25 = 0

Answer: x = 2 ± 1.5 i

BUT an upside-down mirror image of our equation does cross the x-axis at 2 ± 1.5 (note: missing the i ).

Just an interesting fact for you!

• Quadratic Equation in Standard Form: ax 2 + bx + c = 0
• Quadratic Equations can be factored
• Quadratic Formula: x = −b ± √(b 2 − 4ac) 2a
• positive, there are 2 real solutions
• zero, there is one real solution
• negative, there are 2 complex solutions

• Math Article

Quadratics or Quadratic Equations

Quadratics   can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations . The general form of the quadratic equation is: 

ax² + bx + c = 0

where x is an unknown variable and a, b, c are numerical coefficients. For example, x 2 + 2x +1 is a quadratic or quadratic equation. Here, a ≠ 0 because if it equals zero then the equation will not remain quadratic anymore and it will become a linear equation, such as: 

Thus, this equation cannot be called a quadratic equation.

The terms a, b and c are also called quadratic coefficients. 

The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations . The roots of any polynomial are the solutions for the given equation.

What is Quadratic Equation?

The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of:

where x is the unknown variable and a, b and c are the constant terms.

Standard Form of Quadratic Equation

Since the quadratic includes only one unknown term or variable, thus it is called univariate. The power of variable x is always non-negative integers. Hence the equation is a polynomial equation with the highest power as 2.

The solution for this equation is the values of x, which are also called zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x on the left-hand side of the equation, it will equal to zero. Therefore, they are called zeros.

Quadratics Formula

The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:

x = [-b±√(b 2 -4ac)]/2a

The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here.

Examples of Quadratics

Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)

• x² –x – 9 = 0
• 5x² – 2x – 6 = 0
• 3x² + 4x + 8 = 0
• -x² +6x + 12 = 0

Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.

• -x² – 9x = 0
• x² + 2x = 0
• -6x² – 3x = 0
• -5x² + x = 0
• -12x² + 13x = 0
• 11x² – 27x = 0

Following are the examples of a quadratic equation in factored form

• (x – 6)(x + 1) = 0  [ result obtained after solving is x² – 5x – 6 = 0]
• –3(x – 4)(2x + 3) = 0  [result obtained after solving is -6x² + 15x + 36 = 0]
• (x − 5)(x + 3) = 0  [result obtained after solving is x² − 2x − 15 = 0]
• (x – 5)(x + 2) = 0  [ result obtained after solving is x² – 3x – 10 = 0]
• (x – 4)(x + 2) = 0  [result obtained after solving is x² – 2x – 8 = 0]
• (2x+3)(3x – 2) = 0  [result obtained after solving is 6x² + 5x – 6]

Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’

• 2x² – 64 = 0
• x² – 16 = 0
• 9x² + 49 = 0
• -2x² – 4 = 0
• 4x² + 81 = 0
• -x² – 9 = 0

How to Solve Quadratic Equations?

There are basically four methods of solving quadratic equations. They are:

• Completing the square

Using Quadratic Formula

• Taking the square root

Factoring of Quadratics

• Begin with a equation of the form ax² + bx + c = 0
• Ensure that it is set to adequate zero.
• Factor the left-hand side of the equation by assuming zero on the right-hand side of the equation.
• Assign each factor equal to zero.
• Now solve the equation in order to determine the values of x.

Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.

(2x+3)(x-2)=0

Learn more about the factorization of quadratic equations here.

Completing the Square Method

Let us learn this method with example.

Example: Solve 2x 2 – x – 1 = 0.

First, move the constant term to the other side of the equation.

2x 2 – x = 1

Dividing both sides by 2.

x 2 – x/2 = ½

Add the square of half of the coefficient of x, (b/2a) 2 , on both the sides, i.e., 1/16

x 2 – x/2 + 1/16 = ½ + 1/16

Now we can factor the right side,

(x-¼) 2 = 9/16 = (¾) 2

Taking root on both sides;

X – ¼ = ±3/4

Add ¼ on both sides

X = ¼ + ¾ = 4/4 = 1

X = ¼ – ¾ = -2/4 = -½ 

To learn more about completing the square method, click here .

For the given Quadratic equation of the form, ax² + bx + c = 0

Therefore the roots of the given equation can be found by:

\(\begin{array}{l}x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\end{array} \)

where ± (one plus and one minus) represent two distinct roots of the given equation.

Taking the Square Root

We can use this method for the equations such as:

x 2 + a 2 = 0

Example: Solve x 2 – 50 = 0.

x 2 – 50 = 0

Taking the roots both sides

√x 2 = ±√50

x = ±√(2 x 5 x 5)

Thus, we got the required solution.

Related Articles

Video lesson on quadratic equations, range of quadratic equations.

Solved Problems on Quadratic Equations

Applications of quadratic equations.

Many real-life word problems can be solved using quadratic equations. While solving word problems, some common quadratic equation applications include speed problems and Geometry area problems.

• Solving the problems related to finding the area of quadrilateral such as rectangle, parallelogram and so on
• Solving Word Problems involving Distance, speed, and time, etc.,

Example: Find the width of a rectangle of area 336 cm2 if its length is equal to the 4 more than twice its width. Solution: Let x cm be the width of the rectangle. Length = (2x + 4) cm We know that Area of rectangle = Length x Width x(2x + 4) = 336 2x 2 + 4x – 336 = 0 x 2 + 2x – 168 = 0 x 2 + 14x – 12x – 168 = 0 x(x + 14) – 12(x + 14) = 0 (x + 14)(x – 12) = 0 x = -14, x = 12 Measurement cannot be negative. Therefore, Width of the rectangle = x = 12 cm

Practice Questions

• Solve x 2 + 2 x + 1 = 0.
• Solve 5x 2 + 6x + 1 = 0
• Solve 2x 2 + 3 x + 2 = 0.
• Solve x 2 − 4x + 6.25 = 0

Frequently Asked Questions on Quadratics

What is a quadratic equation, what are the methods to solve a quadratic equation, is x 2 – 1 a quadratic equation, what is the solution of x 2 + 4 = 0, write the quadratic equation in the form of sum and product of roots., leave a comment cancel reply.

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Thanks a lot ,This was very useful for me

x=√9 Squaring both the sides, x^2 = 9 x^2 – 9 = 0 It is a quadratic equation.

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Forms of Quadratics: Explanations, Tips, and Examples

• The Albert Team
• Last Updated On: March 1, 2022

Looking to understand the different forms of quadratic equations? Read below for an explanation of the three main forms of quadratics (standard form, factored form, and vertex form), examples of each form, as well as strategies for converting between the various quadratic forms.

Your mathematics journey has taken you far. There was a time when the words “variable” and “equation” were only concepts you would someday understand. The skills you developed then gave you a foundation of using mathematics to solve simple problems. Then, you delved into variables and unknown values. You conquered solving equations for the value of x . You are able to create and interpret graphs of equations.

Now, we are opening a new tool: quadratics ! Quadratic equations may feel different, scary, exciting, or all of the above. Regardless of how you feel going into learning quadratic equations, know that you can conquer this, too. You are entering a new level of mathematical understanding and a new world of real-life situations to model. Let’s get started!

What We Review

The 3 Forms of Quadratic Equations

There are three commonly-used forms of quadratics:

Each quadratic form looks unique, allowing for different problems to be more easily solved in one form than another. We will unpack the features of each form and how to switch between forms.

Return to the Table of Contents

Why are There Forms of Quadratic Equations?

Each form of a quadratic equation includes specific advantages. Recognizing the benefits of each different form can make it easier to understand and solve different situations.

What Does the Standard Form of a Quadratic Tell You?

Let us begin with the benefits of standard form. In standard mathematical notation, formulas and equations are written with the highest degree first. The degree refers to the exponent. In the case of quadratic equations, the degree is two because the highest exponent is two. Following the x^2 term is the term with an exponent of one followed by the term with an exponent of zero.

The benefits of standard form include quickly identifying the end behavior of a function and identifying the values of a, b, and c .

The end behavior of a function is identified by the leading coefficient and the degree of a function. The degree of a quadratic equation is always two. The leading coefficient of a quadratic equation is always the term a when written in standard form.

If the value of a is positive, the parabola opens up, meaning the function rises to the left and rises to the right. If the value of a is negative, the parabola opens down, meaning the function falls to the left and falls to the right.

y=3x^2+2x-1

Notice : positive a value, degree of 2 , parabola opens “up”

y=-3x^2+2x+1

Notice : negative a value, degree of 2 , parabola opens “down”

One method for solving a quadratic equation is to use the quadratic formula . To do so, we must identify the values of a, b, and c . To learn more about this, read our detailed review article on the quadratic formula .

What Does the Factored Form of a Quadratic Tell You?

Next, let’s now consider why factored form is useful. To get to factored form, we do exactly what it sounds like: we factor the equation from standard form.

In the factored form of a quadratic, we are also able to determine end behavior using the value of a . Although the degree is not as easily identifiable, we know there are only two factors, making the degree two. The end behavior follows the same rules explained above.

The additional benefit of factored form is identifying zeros, or x -intercepts, of the function. The value of r_1 and the value of r_2 are both zeros (also called “solutions”) of the quadratic function.

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

We must note that not all quadratics have “real” zeros (some quadratics require imaginary numbers as their zeros), so factored form may not always be applicable.

What Does the Vertex Form of a Quadratic Tell You?

Finally, we have the vertex form of a quadratic. Remember, the vertex is the point where the axis of symmetry intersects the parabola. It is also the lowest point of a parabola opening up or the highest point of a parabola opening down.

As you may expect, the main benefit of vertex form is easily identifying the vertex. The vertex of a parabola, or a quadratic equation, is written as (h,k) where the h is the x -coordinate and the k is the y -coordinate.

As we can see, the value of h and the value of k are easily identifiable in this form. Additionally, we can still determine the end behavior using the value of a .

y= 3(x-2)^2 - 1

What is the Standard Form of a Quadratic?

Remember, standard form provides us values for the coefficients a, b , and c , while x and y are the variables.

For example, the equation:

…is in standard form, telling us that a=3, b=7, and c=-9 . 

What is the Factored Form of a Quadratic?

In factored form , we can see the zeros, also called x -intercepts, are r_1 and r_2 . Our variables remain x and y , and a is a coefficient.

In the factored form equation:

…the two x -intercepts are -8 and 6 , and the value of a equals 3 .

What is the Vertex Form of a Quadratic?

In vertex form , the variables x and y and the coefficient of a still remain, but now we can identify the vertex using the values of h and k .

For instance, in the equation:

…which is in vertex form, the vertex is (2,16) and the value of a is -2 .

Converting Between Forms of Quadratic Equations

Often, we need many different pieces of information about quadratic equations. It can be useful to see the same quadratic equation in the multiple forms. Just like a chameleon can change colors in different situations, we can change the forms of quadratics to suit our needs.

Change forms of a quadratic like a chameleon!

The ability to switch between forms quickly and accurately enables us to understand the quadratic equation well and easily identify needed pieces of information. For example, you may be asked to determine the zeros of a quadratic equation given in standard form. In order to identify the zeros, we first must change the equation to factored form.

Convert from Standard Form to Factored Form

Let us begin with the equation:

y=x^2+5x-24

We may be asked for the zeros of the equation. In order to determine the zeros, we can change this into factored form. To change this into factored form, we must factor the expression x^2+5x-24 . Let’s remember what Factored Form looks like:

In order to factor the expression, we must determine the factors of -24 that have a sum of 5 . We can list the factors of -24 and their sums:

The two values that multiply to -24 and have a sum of 5 are -3 and 8 . Therefore, we can rewrite our quadratic equation by factoring.

To determine the zeros, we set the equation equal to zero . Then, we can solve by setting each factor equal to zero:

Therefore, the zeros of the function are 3 and -8 . The final factored form the equation is:

y=(x-3)(x+8)

To learn more about this approach by reading our article on solving quadratic equations by factoring.

Convert from Standard Form to Vertex Form

Instead of being asked for the zeros, we could be asked for the vertex of a quadratic equation. Let us begin with the quadratic equation:

…which is given in standard form, and determine the vertex of the equation. In order to do so, we will convert this into vertex form.

To convert into vertex form, we must complete a process called “completing the square”. Essentially, we are setting up a trinomial that we can factor into a perfect square. Let us remember what vertex form of a quadratic looks like:

We need to set up the equation just right so that we can factor it to create (x-h)^2 . This may sound intimidating, but there is a step-by-step process that always works!

First, isolate the x . This means we need to move any constants to the side with y . Constants are terms with no variable attached.

If we had a leading coefficient other than one, we would divide all terms by the leading coefficient. Remember, the leading coefficient is the number in front of x^2 . In our case, we have a leading coefficient of one, so we can skip this step.

From here, we need to determine what value to add to both sides. To determine this value, we look at the number in front of x . In our case, this value is 6 . We have half of this value and then square the result.

The value we must add to both sides of our equation is 9 . Remember this will create a trinomial which is a perfect square (thus, the name “completing the square”).

Now we have created a trinomial, x^2+6x+9 , which we can factor into a perfect square. This factors into (x+3)^2 . Notice this matches the step where we took half of 6 . After we factor it, we will then solve for y .

Now that the equation is in vertex form, we can identify the vertex as (-3,-14) . The vertex form of the quadratic equation is:

Converting from quadratic form to standard form is quite common, so you can also check out this helpful video for another example.

Convert from Factored Form to Standard Form

To convert an equation from factored form into standard form simply involves multiplying the factors. For example, let us change the quadratic equation:

y=(3x-2)(-x+7)

…into standard form. We will use double distribution to multiply the factors (3x-2) and (-x+7) together.

Therefore, the standard form of the quadratic equation is y=-3x^2+23x-14 .

Convert from Vertex Form to Standard Form

Finally, we may also need to convert an equation from vertex form into standard form. For example, we can change the equation:

y=2(x+7)^2-10

…into standard form. We will expand the expression (x+7)^2 and again use double distribution. Then, we will continue simplifying the equation.

Therefore, the standard form of the quadratic equation is y=2x^2+28x+88 .

Practice with Forms of Quadratics

For some practice questions covering the forms of quadratics, check out Albert’s Algebra 1 practice course ! All of Albert’s questions include explanations of solutions and how to avoid common mistakes.

Additionally, licensed Albert teachers can assign students this short Algebra 1 Topic Quiz that focuses on vertex, roots, and the various forms of quadratics.

Finally, check out  our other detailed Algebra 1 review guides  to learn more about quadratic.

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5.8: Solve Applications of Quadratic Equations

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Learning Objectives

By the end of this section, you will be able to:

• Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

• The sum of two consecutive odd numbers is \(−100\). Find the numbers. If you missed this problem, review Example 2.18.
• Solve: \(\frac{2}{x+1}+\frac{1}{x-1}=\frac{1}{x^{2}-1}\). If you missed this problem, review Example 7.35.
• Find the length of the hypotenuse of a right triangle with legs \(5\) inches and \(12\) inches. If you missed this problem, review Example 2.34.

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

• Square Root Property
• Completing the Square
• Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy

• Read the problem. Make sure all the words and ideas are understood.
• Identify what we are looking for.
• Name what we are looking for. Choose a variable to represent that quantity.
• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Solve the equation using algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is \(2\) more than the number preceding it. If we call the first one \(n\), then the next one is \(n+2\). The next one would be \(n+2+2\) or \(n+4\). This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

\(\begin{array}{cl}{}&{\text{Consecutive even integers}}\\{}& {64,66,68}\\ {n} & {1^{\text { st }} \text { even integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive even integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive even integer }}\end{array}\)

\(\begin{array}{cl}{}&{\text{Consecutive odd integers}}\\{}& {77,79,81}\\ {n} & {1^{\text { st }} \text { odd integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive odd integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive odd integer }}\end{array}\)

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

Example \(\PageIndex{1}\)

The product of two consecutive odd integers is \(195\). Find the integers.

Step 1 : Read the problem

Step 2 : Identify what we are looking for.

We are looking for two consecutive odd integers.

Step 3 : Name what we are looking for.

Let \(n=\) the first odd integer.

\(n+2=\) the next odd integer.

Step 4 : Translate into an equation. State the problem in one sentence.

“The product of two consecutive odd integers is \(195\).” The product of the first odd integer and the second odd integer is \(195\).

Translate into an equation.

\(n(n+2)=195\)

Step 5 : Solve the equation. Distribute.

\(n^{2}+2 n=195\)

Write the equation in standard form.

\(n^{2}+2 n-195=0\)

\((n+15)(n-13)=0\)

Use the Zero Product Property.

\(n+15=0 \quad n-13=0\)

Solve each equation.

\(n=-15, \quad n=13\)

There are two values of \(n\) that are solutions. This will give us two pairs of consecutive odd integers for our solution.

\(\begin{array}{cc}{\text { First odd integer } n=13} & {\text { First odd integer } n=-15} \\ {\text { next odd integer } n+2} & {\text { next odd integer } n+2} \\ {13+2} & {-15+2} \\ {15} & {-13}\end{array}\)

Step 6 : Check the answer.

Do these pairs work? Are they consecutive odd integers?

\(\begin{aligned} 13,15 & \text { yes } \\-13,-15 & \text { yes } \end{aligned}\)

Is their product \(195\)?

\(\begin{aligned} 13 \cdot 15 &=195 &\text{yes} \\-13(-15) &=195 & \text { yes } \end{aligned}\)

Step 7 : Answer the question.

Two consecutive odd integers whose product is \(195\) are \(13,15\) and \(-13,-15\).

Exercise \(\PageIndex{1}\)

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

The two consecutive odd integers whose product is \(99\) are \(9, 11\), and \(−9, −11\).

Exercise \(\PageIndex{2}\)

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

The two consecutive even integers whose product is \(128\) are \(12, 14\) and \(−12, −14\).

We will use the formula for the area of a triangle to solve the next example.

Definition \(\PageIndex{1}\)

Area of a Triangle

For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2} b h\).

Recall that when we solve geometric applications, it is helpful to draw the figure.

Example \(\PageIndex{2}\)

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of \(120\) square feet and the architect wants the base to be \(4\) feet more than twice the height. Find the base and height of the window.

Exercise \(\PageIndex{3}\)

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of \(456\) square inches.

The height of the triangle is \(12\) inches and the base is \(76\) inches.

Exercise \(\PageIndex{4}\)

If a triangle that has an area of \(110\) square feet has a base that is two feet less than twice the height, what is the length of its base and height?

The height of the triangle is \(11\) feet and the base is \(20\) feet.

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Definition \(\PageIndex{2}\)

Area of a Rectangle

For a rectangle with length, \(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

Example \(\PageIndex{3}\)

Mike wants to put \(150\) square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than \(3\) times the width. Find the length and width. Round to the nearest tenth of a foot.

Exercise \(\PageIndex{5}\)

The length of a \(200\) square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

The length of the garden is approximately \(18\) feet and the width \(11\) feet.

Exercise \(\PageIndex{6}\)

A rectangular tablecloth has an area of \(80\) square feet. The width is \(5\) feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot?

The length of the tablecloth is approximately \(11.8\) feet and the width \(6.8\) feet.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Definition \(\PageIndex{3}\)

Pythagorean Theorem

• In any right triangle, where \(a\) and \(b\) are the lengths of the legs, and \(c\) is the length of the hypotenuse, \(a^{2}+b^{2}=c^{2}\).

Example \(\PageIndex{4}\)

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two \(10\)-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Exercise \(\PageIndex{7}\)

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is \(20\) feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

The length of the flag pole’s shadow is approximately \(6.3\) feet and the height of the flag pole is \(18.9\) feet.

Exercise \(\PageIndex{8}\)

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The distance between the opposite corners is approximately \(7.2\) feet.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, \(v_{0}\), propels the object up until gravity causes the object to fall back down.

Definition \(\PageIndex{4}\)

The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula

\(h=-16 t^{2}+v_{0} t\)

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

Example \(\PageIndex{5}\)

A firework is shot upwards with initial velocity \(130\) feet per second. How many seconds will it take to reach a height of \(260\) feet? Round to the nearest tenth of a second.

Exercise \(\PageIndex{9}\)

An arrow is shot from the ground into the air at an initial speed of \(108\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the arrow will be \(180\) feet from the ground. Round the nearest tenth.

The arrow will reach \(180\) feet on its way up after \(3\) seconds and again on its way down after approximately \(3.8\) seconds.

Exercise \(\PageIndex{10}\)

A man throws a ball into the air with a velocity of \(96\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the height of the ball will be \(48\) feet. Round to the nearest tenth.

The ball will reach \(48\) feet on its way up after approximately \(.6\) second and again on its way down after approximately \(5.4\) seconds.

We have solved uniform motion problems using the formula \(D=rt\) in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The formula \(D=rt\) assumes we know \(r\) and \(t\) and use them to find \(D\). If we know \(D\) and \(r\) and need to find \(t\), we would solve the equation for \(t\) and get the formula \(t=\frac{D}{r}\).

Some uniform motion problems are also modeled by quadratic equations.

Example \(\PageIndex{6}\)

Professor Smith just returned from a conference that was \(2,000\) miles east of his home. His total time in the airplane for the round trip was \(9\) hours. If the plane was flying at a rate of \(450\) miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.

We fill in the chart to organize the information.

We are looking for the speed of the jet stream. Let \(r=\) the speed of the jet stream.

When the plane flies with the wind, the wind increases its speed and so the rate is \(450 + r\).

When the plane flies against the wind, the wind decreases its speed and the rate is \(450 − r\).

The speed of the jet stream was \(50\) mph.

Exercise \(\PageIndex{11}\)

MaryAnne just returned from a visit with her grandchildren back east. The trip was \(2400\) miles from her home and her total time in the airplane for the round trip was \(10\) hours. If the plane was flying at a rate of \(500\) miles per hour, what was the speed of the jet stream?

The speed of the jet stream was \(100\) mph.

Exercise \(\PageIndex{12}\)

Gerry just returned from a cross country trip. The trip was \(3000\) miles from his home and his total time in the airplane for the round trip was \(11\) hours. If the plane was flying at a rate of \(550\) miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

Example \(\PageIndex{7}\)

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(12\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(8\) hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Exercise \(\PageIndex{13}\)

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(6\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(4\) hours. How long does it take for each press to print the job alone?

Press #1 would take \(12\) hours, and Press #2 would take \(6\) hours to do the job alone.

Exercise \(\PageIndex{14}\)

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes \(3\) hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in \(2\) hours. How long does it take for each hose to fill the hot tub?

The red hose take \(6\) hours and the green hose take \(3\) hours alone.

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

• Word Problems Involving Quadratic Equations
• Quadratic Equation Word Problems
• Applying the Quadratic Formula

Key Concepts

• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
• Solve the equation using good algebra techniques.
• For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2}bh\).

• For a rectangle with length,\(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

• The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula \(h=-16 t^{2}+v_{0} t\).