Contents

Quadratic Equations

Learning Objectives

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

Solve quadratic equations by using the Zero Product Property
Solve quadratic equations factoring
Solve applications modeled by quadratic equations

Be Prepared 7.6

Before you get started, take this readiness quiz.

Solve: $5 y - 3 = 0 5 y - 3 = 0$ .
If you missed this problem, review Example 2.27.
Solve: $10 a = 0 10 a = 0$ .
If you missed this problem, review Example 2.13.
Combine like terms: $12 x^{2} - 6 x + 4 x 12 x^{2} - 6 x + 4 x$ .
If you missed this problem, review Example 1.24.
Factor $n^{3} - 9 n^{2} - 22 n n^{3} - 9 n^{2} - 22 n$ completely.
If you missed this problem, review Example 7.32.

We have already solved linear equations, equations of the form $a x + b y = c a x + b y = c$ . In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. Listed below are some examples of quadratic equations:

\begin{matrix} x^{2} + 5 x + 6 = 0 & 3 y^{2} + 4 y = 10 & 64 u^{2} - 81 = 0 & n (n + 1) = 42 \end{matrix} \begin{matrix} x^{2} + 5 x + 6 = 0 & 3 y^{2} + 4 y = 10 64 u^{2} - 81 = 0 n (n + 1) = 42 \end{matrix}

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get $n^{2} + n n^{2} + n$ .

The general form of a quadratic equation is $a x^{2} + b x + c = 0, with a \neq 0 a x^{2} + b x + c = 0, with$ .

Quadratic Equation

An equation of the form $a x^{2} + b x + c = 0 a x^{2} + b x + c = 0$ is called a quadratic equation.

a, b, and c are real numbers and a \neq 0 a, b,

To solve quadratic equations we need methods different than the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

Solve Quadratic Equations Using the Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Zero Product Property

If $a \cdot b = 0 a \cdot b = 0$ , then either $a = 0 a = 0$ or $b = 0 b = 0$ or both.

We will now use the Zero Product Property, to solve a quadratic equation.

Example 7.69

How to Use the Zero Product Property to Solve a Quadratic Equation

Solve: $(x + 1) (x - 4) = 0 (x + 1) (x - 4) = 0$ .

Show solution

Solution

This table gives the steps for solving (x + 1)(x – 4) = 0. The first step is to set each factor equal to 0. Since it is a product equal to 0, at least one factor must equal 0. x + 1 = 0 or x – 4 = 0.

The last step is to check both answers by substituting the values for x into the original equation.

Try It 7.137

Solve: $(x - 3) (x + 5) = 0 (x - 3) (x + 5) = 0$ .

Try It 7.138

Solve: $(y - 6) (y + 9) = 0 (y - 6) (y + 9) = 0$ .

We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property.

Example 7.70

Solve: $(5 n - 2) (6 n - 1) = 0 (5 n - 2) (6 n - 1) = 0$ .

Show solution

Solution

	$(5 n - 2) (6 n - 1) = 0 (5 n - 2) (6 n - 1) = 0$
Use the Zero Product Property to set each factor to 0.	$5 n - 2 = 0 5 n - 2 = 0$	$6 n - 1 = 0 6 n - 1 = 0$
Solve the equations.	$n = \frac{2}{5} n = \frac{2}{5}$	$n = \frac{1}{6} n = \frac{1}{6}$
Check your answers.

Try It 7.139

Solve: $(3 m - 2) (2 m + 1) = 0 (3 m - 2) (2 m + 1) = 0$ .

Try It 7.140

Solve: $(4 p + 3) (4 p - 3) = 0 (4 p + 3) (4 p - 3) = 0$ .

Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.

Example 7.71

Solve: $3 p (10 p + 7) = 0 3 p (10 p + 7) = 0$ .

Show solution

Solution

	$3 p (10 p + 7) = 0$
Use the Zero Product Property to set each factor to 0.	$3 p = 0 3 p = 0$	$10 p + 7 = 0 10 p + 7 = 0$
Solve the equations.	$p = 0 p = 0$	$10 p = −7 10 p = −7$
		$p = - \frac{7}{10} p = - \frac{7}{10}$
Check your answers.

Try It 7.141

Solve: $2 u (5 u - 1) = 0 2 u (5 u - 1) = 0$ .

Try It 7.142

Solve: $w (2 w + 3) = 0 w (2 w + 3) = 0$ .

It may appear that there is only one factor in the next example. Remember, however, that ${(y - 8)}^{2} {(y - 8)}^{2}$ means $(y - 8) (y - 8) (y - 8) (y - 8)$ .

Example 7.72

Solve: ${(y - 8)}^{2} = 0 {(y - 8)}^{2} = 0$ .

Show solution

Solution

	${(y - 8)}^{2} = 0 {(y - 8)}^{2} = 0$
Rewrite the left side as a product.	$(y - 8) (y - 8) = 0 (y - 8) (y - 8) = 0$
Use the Zero Product Property and set each factor to 0.	$y - 8 = 0 y - 8 = 0$	$y - 8 = 0 y - 8 = 0$
Solve the equations.	$y = 8 y = 8$	$y = 8 y = 8$
When a solution repeats, we call it a double root.
Check your answer.

Try It 7.143

Solve: ${(x + 1)}^{2} = 0 {(x + 1)}^{2} = 0$ .

Try It 7.144

Solve: ${(v - 2)}^{2} = 0 {(v - 2)}^{2} = 0$ .

Solve Quadratic Equations by Factoring

Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we be sure to start with the quadratic equation in standard form, $a x^{2} + b x + c = 0 a x^{2} + b x + c = 0$ . Then we factor the expression on the left.

Example 7.73

How to Solve a Quadratic Equation by Factoring

Solve: $x^{2} + 2 x - 8 = 0 x^{2} + 2 x - 8 = 0$ .

Show solution

Solution

This table gives the steps for solving the equation x squared + 2 x – 8 = 0. The first step is writing the equation in standard quadratic form, which it is.

The second step is factoring the quadratic expression x squared + 2 x – 8. The factors are (x + 4), (x – 2).

The last step is checking both solutions by substituting them into the original equation.

Try It 7.145

Solve: $x^{2} - x - 12 = 0 x^{2} - x - 12 = 0$ .

Try It 7.146

Solve: $b^{2} + 9 b + 14 = 0 b^{2} + 9 b + 14 = 0$ .

How To

Solve a quadratic equation by factoring.

Step 1. Write the quadratic equation in standard form, $a x^{2} + b x + c = 0 a x^{2} + b x + c = 0$ .
Step 2. Factor the quadratic expression.
Step 3. Use the Zero Product Property.
Step 4. Solve the linear equations.
Step 5. Check.

Before we factor, we must make sure the quadratic equation is in standard form.

Example 7.74

Solve: $2 y^{2} = 13 y + 45 2 y^{2} = 13 y + 45$ .

Show solution

Solution

	$2 y^{2} = 13 y + 45$
Write the quadratic equation in standard form.	$2 y^{2} - 13 y - 45 = 0 2 y^{2} - 13 y - 45 = 0$
Factor the quadratic expression.	$(2 y + 5) (y - 9) = 0$
Use the Zero Product Property to set each factor to 0.	$2 y + 5 = 0$	$y - 9 = 0 y - 9 = 0$
Solve each equation.	$y = - \frac{5}{2}$	$y = 9 y = 9$
Check your answers.

Try It 7.147

Solve: $3 c^{2} = 10 c - 8 3 c^{2} = 10 c - 8$ .

Try It 7.148

Solve: $2 d^{2} - 5 d = 3 2 d^{2} - 5 d = 3$ .

Example 7.75

Solve: $5 x^{2} - 13 x = 7 x 5 x^{2} - 13 x = 7 x$ .

Show solution

Solution

	$5 x^{2} - 13 x = 7 x 5 x^{2} - 13 x = 7 x$
Write the quadratic equation in standard form.	$5 x^{2} - 20 x = 0 5 x^{2} - 20 x = 0$
Factor the left side of the equation.	$5 x (x - 4) = 0 5 x (x - 4) = 0$
Use the Zero Product Property to set each factor to 0.	$5 x = 0 5 x = 0$	$x - 4 = 0 x - 4 = 0$
Solve each equation.	$x = 0 x = 0$	$x = 4 x = 4$
Check your answers.

Try It 7.149

Solve: $6 a^{2} + 9 a = 3 a 6 a^{2} + 9 a = 3 a$ .

Try It 7.150

Solve: $45 b^{2} - 2 b = −17 b 45 b^{2} - 2 b = −17 b$ .

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Example 7.76

Solve: $144 q^{2} = 25 144 q^{2} = 25$ .

Show solution

Solution


Write the quadratic equation in standard form.
Factor. It is a difference of squares.
Use the Zero Product Property to set each factor to 0.	$\begin{matrix} 12 q - 5 = 0 \end{matrix} \begin{matrix} 12 q - 5 = 0 \end{matrix}$	$\begin{matrix} 12 q + 5 = 0 \end{matrix} \begin{matrix} 12 q + 5 = 0 \end{matrix}$
Solve each equation.	$\begin{matrix} 12 q & = & 5 \\ q & = & \frac{5}{12} \end{matrix} \begin{matrix} 12 q & = & 5 \\ q & = & \frac{5}{12} \end{matrix}$	$\begin{matrix} 12 q & = & –5 \\ q & = & - \frac{5}{12} \end{matrix} \begin{matrix} 12 q & = & –5 \\ q & = & - \frac{5}{12} \end{matrix}$
Check your answers.

Try It 7.151

Solve: $25 p^{2} = 49 25 p^{2} = 49$ .

Try It 7.152

Solve: $36 x^{2} = 121 36 x^{2} = 121$ .

The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Example 7.77

Solve: $(3 x - 8) (x - 1) = 3 x (3 x - 8) (x - 1) = 3 x$ .

Show solution

Solution

$\begin{matrix} \begin{matrix} Multiply the binomials. \\ Write the quadratic equation in standard form. \\ Factor the trinomial. \end{matrix} & \begin{matrix} (3 x - 8) (x - 1) & = & 3 x \\ 3 x^{2} - 11 x + 8 & = & 3 x \\ 3 x^{2} - 14 x + 8 & = & 0 \\ (3 x - 2) (x - 4) & = & 0 \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0. \\ Solve each equation. \end{matrix} & \begin{matrix} 3 x - 2 & = & 0 & x - 4 & = & 0 \\ 3 x & = & 2 & x & = & 4 \\ x & = & \frac{2}{3} \end{matrix} \\ Check your answers. & The check is left to you! \end{matrix} \begin{matrix} \begin{matrix} Multiply the binomials. & Write the quadratic equation in standard form. & Factor the trinomial. \end{matrix} & \begin{matrix} (3 x - 8) (x - 1) & = & 3 x \\ 3 x^{2} - 11 x + 8 & = & 3 x & 3 x^{2} - 14 x + 8 & = & 0 & (3 x - 2) (x - 4) & = & 0 \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0. \\ Solve each equation. \end{matrix} & Check your answers. & The check is left to you! \end{matrix}$

Try It 7.153

Solve: $(2 m + 1) (m + 3) = 12 m (2 m + 1) (m + 3) = 12 m$ .

Try It 7.154

Solve: $(k + 1) (k - 1) = 8 (k + 1) (k - 1) = 8$ .

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations.

Example 7.78

Solve: $9 m^{3} + 100 m = 60 m^{2} 9 m^{3} + 100 m = 60 m^{2}$ .

Show solution

Solution


Bring all the terms to one side so that the other side is zero.
Factor the greatest common factor first.
Factor the trinomial.
Use the Zero Product Property to set each factor to 0.
Solve each equation.
Check your answers.	The check is left to you.

Try It 7.155

Solve: $8 x^{3} = 24 x^{2} - 18 x 8 x^{3} = 24 x^{2} - 18 x$ .

Try It 7.156

Solve: $16 y^{2} = 32 y^{3} + 2 y 16 y^{2} = 32 y^{3} + 2 y$ .

When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Example 7.79

Solve: $4 x^{2} = 16 x + 84 4 x^{2} = 16 x + 84$ .

Show solution

Solution

$\begin{matrix} \begin{matrix} Write the quadratic equation in standard form. \\ Factor the greatest common factor first. \\ Factor the trinomial. \end{matrix} & \begin{matrix} 4 x^{2} & = & 16 x + 84 \\ 4 x^{2} - 16 x - 84 & = & 0 \\ 4 (x^{2} - 4 x - 21) & = & 0 \\ 4 (x - 7) (x + 3) & = & 0 \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0. \\ Solve each equation. \end{matrix} & \begin{matrix} 4 & \neq & 0 & x - 7 & = & 0 & x + 3 & = & 0 \\ 4 & \neq & 0 & x & = & 7 & x & = & −3 \end{matrix} \\ Check your answers. & The check is left to you. \end{matrix} \begin{matrix} \begin{matrix} Write the quadratic equation in standard form. & Factor the greatest common factor first. & Factor the trinomial. \end{matrix} \\ \begin{matrix} Use the Zero Product Property to set each factor to 0. \\ Solve each equation. \end{matrix} & Check your answers. & The check is left to you. \end{matrix}$

Try It 7.157

Solve: $18 a^{2} - 30 = −33 a 18 a^{2} - 30 = −33 a$ .

Try It 7.158

Solve: $123 b = −6 - 60 b^{2} 123 b = −6 - 60 b^{2}$ .

Solve Applications Modeled by Quadratic Equations

The problem solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We will copy the problem solving strategy here so we can use it for reference.

How To

Use a problem-solving strategy to solve word problems.

Step 1. Read the problem. Make sure all the words and ideas are understood.
Step 2. Identify what we are looking for.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
Step 4. 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.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a quadratic equation.

Example 7.80

The product of two consecutive integers is $132132$ . Find the integers.

Show solution

Solution

$\begin{matrix} Step 1. Read the problem. \\ Step 2. Identify what we are looking for. & We are looking for two consecutive integers. \\ Step 3. Name what we are looking for. & Let n = the first integer \\ n + 1 = the next consecutive integer \\ Step 4. Translate into an equation. Restate the & The product of the two consecutive integers is 132. \\ problem in a sentence. \\ The first integer times the next integer is 132. \\ \begin{matrix} Translate to an equation. \\ Step 5. Solve the equation. \\ Bring all the terms to one side. \\ Factor the trinomial. \end{matrix} & \begin{matrix} n (n + 1) & = & 132 \\ n^{2} + n & = & 132 \\ n^{2} + n - 132 & = & 0 \\ (n - 11) (n + 12) & = & 0 \end{matrix} \\ \begin{matrix} Use the zero product property. \\ Solve the equations. \end{matrix} & \begin{matrix} n - 11 & = & 0 & n + 12 & = & 0 \\ n & = & 11 & n & = & −12 \end{matrix} \end{matrix} \begin{matrix} Step 1. Read \\ Step 2. Identify & We are looking for two consecutive integers. & Step 3. Name & Let & n + 1 = & Step 4. Translate & The product of the two consecutive integers is 132. & problem in a sentence. & The first integer times the next integer is 132. & \begin{matrix} Translate to an equation. \\ Step 5. Solve & Bring all the terms to one side. & Factor the trinomial. \end{matrix} & \begin{matrix} n (n + 1) & = & 132 \\ n^{2} + n & = & 132 & n^{2} + n - 132 & = & 0 & (n - 11) (n + 12) & = & 0 \end{matrix} & \begin{matrix} Use the zero product property. \\ Solve the equations. \end{matrix} \end{matrix}$

There are two values for $n n$ that are solutions to this problem. So there are two sets of consecutive integers that will work.

$\begin{matrix} If the first integer is n = 11 & If the first integer is n = −12 \\ then the next integer is n + 1 & then the next integer is n + 1 \\ 11 + 1 & −12 + 1 \\ 12 & −11 \end{matrix} \begin{matrix} If the first integer is & If the first integer is \\ then the next integer is & then the next integer is & 11 + 1 & −12 + 1 & 12 & −11 \end{matrix}$

Step 6. Check the answer.

The consecutive integers are $11, 12 11, 12$ and $−11, −12 −11, −12$ . The product $11 \cdot 12 = 132 11 \cdot 12 = 132$ and the product $−11 (−12) = 132 −11 (−12) = 132$ . Both pairs of consecutive integers are solutions.

Step 7. Answer the question. The consecutive integers are $11, 12 11, 12$ and $−11, −12 −11, −12$ .

Try It 7.159

The product of two consecutive integers is $240240$ . Find the integers.

Try It 7.160

The product of two consecutive integers is $420420$ . Find the integers.

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give 132.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

Example 7.81

A rectangular garden has an area $1515$ square feet. The length of the garden is two feet more than the width. Find the length and width of the garden.

Show solution

Solution

Step 1. Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation.
Step 2. Identify what you are looking for.	We are looking for the length and width.
Step 3. Name what you are looking for. The length is two feet more than width.	Let W = the width of the garden. W + 2 = the length of the garden
Step 4. Translate into an equation. Restate the important information in a sentence.	The area of the rectangular garden is 15 square feet.
Use the formula for the area of a rectangle.	$A = L \cdot W$
Substitute in the variables.	$15 = (W + 2) W$
Step 5. Solve the equation. Distribute first.	$15 = W^{2} + 2 W$
Get zero on one side.	$0 = W^{2} + 2 W - 15$
Factor the trinomial.	$0 = (W + 5) (W - 3)$
Use the Zero Product Property.	$0 = W + 5$	$0 = W - 3 0 = W - 3$
Solve each equation.	$−5 = W$	$3 = W 3 = W$
Since W is the width of the garden, it does not make sense for it to be negative. We eliminate that value for W.	$−5 = W$ $W = 3$	$3 = W 3 = W$ Width is 3 feet.
Find the value of the length.	$W + 2 = length$
	$3 + 2$
	$5$	Length is 5 feet.
Step 6. Check the answer. Does the answer make sense?

	Yes, this makes sense.
Step 7. Answer the question.	The width of the garden is 3 feet and the length is 5 feet.

Try It 7.161

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

Try It 7.162

A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

In an earlier chapter, we used the Pythagorean Theorem $(a^{2} + b^{2} = c^{2}) (a^{2} + b^{2} = c^{2})$ . It gave the relation between the legs and the hypotenuse of a right triangle.

We will use this formula to in the next example.

Example 7.82

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle, as shown below. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the deck.

This figure is a right triangle. The vertical leg is labeled “x – 7”. the horizontal leg, the base, is labeled “x”. The hypotenuse is labeled “17”.

Show solution

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	We are looking for the lengths of the sides of the deck.
Step 3. Name what you are looking for. One side is 7 less than the other.	Let x = length of a side of the deck x − 7 = length of other side
Step 4. Translate into an equation. Since this is a right triangle we can use the Pythagorean Theorem.	$a^{2} + b^{2} = c^{2}$
Substitute in the variables.	$x^{2} + {(x - 7)}^{2} = 17^{2}$
Step 5. Solve the equation.	$x^{2} + x^{2} - 14 x + 49 = 289$
Simplify.	$2 x^{2} - 14 x + 49 = 289$
It is a quadratic equation, so get zero on one side.	$2 x^{2} - 14 x - 240 = 0$
Factor the greatest common factor.	$2 (x^{2} - 7 x - 120) = 0$
Factor the trinomial.	$2 (x - 15) (x + 8) = 0$
Use the Zero Product Property.	$2 \neq 0 2 \neq 0$	$x - 15 = 0$	$x + 8 = 0 x + 8 = 0$
Solve.	$2 \neq 0 2 \neq 0$	$x = 15$	$x = −8$
Since x is a side of the triangle, $x = −8 x = −8$ does not make sense.	$2 \neq 0 2 \neq 0$	$x = 15$	$x = −8$
Find the length of the other side.
If the length of one side is
then the length of the other side is

	8 is the length of the other side.
Step 6. Check the answer. Do these numbers make sense?

Step 7. Answer the question.	The sides of the deck are 8, 15, and 17 feet.

Try It 7.163

A boat’s sail is a right triangle. The length of one side of the sail is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the sail.

Try It 7.164

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg.

Practice Makes Perfect

Use the Zero Product Property

In the following exercises, solve.

315.

$(x - 3) (x + 7) = 0 (x - 3) (x + 7) = 0$

316.

$(y - 11) (y + 1) = 0 (y - 11) (y + 1) = 0$

317.

$(3 a - 10) (2 a - 7) = 0 (3 a - 10) (2 a - 7) = 0$

318.

$(5 b + 1) (6 b + 1) = 0 (5 b + 1) (6 b + 1) = 0$

319.

$6 m (12 m - 5) = 0 6 m (12 m - 5) = 0$

320.

$2 x (6 x - 3) = 0 2 x (6 x - 3) = 0$

321.

${(y - 3)}^{2} = 0 {(y - 3)}^{2} = 0$

322.

${(b + 10)}^{2} = 0 {(b + 10)}^{2} = 0$

323.

${(2 x - 1)}^{2} = 0 {(2 x - 1)}^{2} = 0$

324.

${(3 y + 5)}^{2} = 0 {(3 y + 5)}^{2} = 0$

Solve Quadratic Equations by Factoring

In the following exercises, solve.

325.

$x^{2} + 7 x + 12 = 0 x^{2} + 7 x + 12 = 0$

326.

$y^{2} - 8 y + 15 = 0 y^{2} - 8 y + 15 = 0$

327.

$5 a^{2} - 26 a = 24 5 a^{2} - 26 a = 24$

328.

$4 b^{2} + 7 b = −3 4 b^{2} + 7 b = −3$

329.

$4 m^{2} = 17 m - 15 4 m^{2} = 17 m - 15$

330.

$n^{2} = 5 - 6 n n^{2} = 5 - 6 n$ $n^{2} = 5 n - 6 n^{2} = 5 n - 6$

331.

$7 a^{2} + 14 a = 7 a 7 a^{2} + 14 a = 7 a$

332.

$12 b^{2} - 15 b = −9 b 12 b^{2} - 15 b = −9 b$

333.

$49 m^{2} = 144 49 m^{2} = 144$

334.

$625 = x^{2} 625 = x^{2}$

335.

$(y - 3) (y + 2) = 4 y (y - 3) (y + 2) = 4 y$

336.

$(p - 5) (p + 3) = −7 (p - 5) (p + 3) = −7$

337.

$(2 x + 1) (x - 3) = −4 x (2 x + 1) (x - 3) = −4 x$

338.

$(x + 6) (x - 3) = −8 (x + 6) (x - 3) = −8$

339.

$16 p^{3} = 24 p^{2} + 9 p 16 p^{3} = 24 p^{2} + 9 p$

340.

$m^{3} - 2 m^{2} = − m m^{3} - 2 m^{2} = − m$

341.

$20 x^{2} - 60 x = −45 20 x^{2} - 60 x = −45$

342.

$3 y^{2} - 18 y = −27 3 y^{2} - 18 y = −27$

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

343.

The product of two consecutive integers is 56. Find the integers.

344.

The product of two consecutive integers is 42. Find the integers.

345.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

346.

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

347.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

348.

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

Mixed Practice

In the following exercises, solve.

349.

$(x + 8) (x - 3) = 0 (x + 8) (x - 3) = 0$

350.

$(3 y - 5) (y + 7) = 0 (3 y - 5) (y + 7) = 0$

351.

$p^{2} + 12 p + 11 = 0 p^{2} + 12 p + 11 = 0$

352.

$q^{2} - 12 q - 13 = 0 q^{2} - 12 q - 13 = 0$

353.

$m^{2} = 6 m + 16 m^{2} = 6 m + 16$

354.

$4 n^{2} + 19 n = 5 4 n^{2} + 19 n = 5$

355.

$a^{3} - a^{2} - 42 a = 0 a^{3} - a^{2} - 42 a = 0$

356.

$4 b^{2} - 60 b + 224 = 0 4 b^{2} - 60 b + 224 = 0$

357.

The product of two consecutive integers is 110. Find the integers.

358.

The length of one leg of a right triangle is three more than the other leg. If the hypotenuse is 15, find the lengths of the two legs.

Everyday Math

359.

Area of a patio If each side of a square patio is increased by 4 feet, the area of the patio would be 196 square feet. Solve the equation ${(s + 4)}^{2} = 196 {(s + 4)}^{2} = 196$ for s to find the length of a side of the patio.

360.

Watermelon drop A watermelon is dropped from the tenth story of a building. Solve the equation $−16 t^{2} + 144 = 0 −16 t^{2} + 144 = 0$ for $t t$ to find the number of seconds it takes the watermelon to reach the ground.

Writing Exercises

361.

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

362.

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

Self Check

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

This table has the following statements all to be preceded by “I can…”. The first row is “solve quadratic equations by using the zero product property”. The second row is “solve quadratic equations by factoring”. The third row is “solve applications modeled by quadratic equations”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?