1. Learning objectives

Learning objectives

Solve quadratic equations by using the Zero Product Property

Solve quadratic equations factoring

Solve applications modeled by quadratic equations

We have already solved linear equations, equations of the form $ax+by=c$ . 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{array}{cccccccccc} {x}^{2}+5x+6=0 & & & 3{y}^{2}+4y=10 & & & 64{u}^{2}-81=0 & & & n(n+1)=42 \\ \end{array}

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

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

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

a,b,\,\mathrm{and}\,c\,\mathrm{are real numbers and}\,a\ne{}0

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.

2. Solve Quadratic Equations Using the Zero Product Property

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.

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

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

3. Example 7.69

Example 7.69

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

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

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.

4. Example 7.70

Example 7.70

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

Solution

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.

5. Example 7.71

Example 7.71

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

Solution

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

6. Example 7.72

Example 7.72

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

Solution

7. Solve Quadratic Equations by Factoring

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 must be sure to start with the quadratic equation in standard form, $a{x}^{2}+bx+c=0$ . Then we can factor the expression on the left.

8. Example 7.73

Example 7.73

How to Solve a Quadratic Equation by Factoring

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

Solve a quadratic equation by factoring.

Step 1. Write the quadratic equation in standard form, $a{x}^{2}+bx+c=0$ .
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.

9. Example 7.74

Example 7.74

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

Solution

10. Example 7.75

Example 7.75

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

Solution

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?

11. Example 7.76

Example 7.76

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

Solution

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.

12. Example 7.77

Example 7.77

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

Solution


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

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.

13. Example 7.78

Example 7.78

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

Solution

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.

14. Example 7.79

Example 7.79

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

Solution


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

15. Solve Applications Modeled by Quadratic Equations

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.

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.

16. Example 7.80

Example 7.80

The product of two consecutive integers is 132. Find the integers.

Solution


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.	$\begin{array}{l} \mathrm{Let}\,n=\mathrm{the first integer} \\ n+1=\,\mathrm{the next consecutive integer} \\ \end{array}$
Step 4. Translate into an equation. Restate the problem in a sentence.	The product of the two consecutive integers is 132.
	The first integer times the next integer is 132.
Translate to an equation.	$\,\begin{array}{rll} n(n+1) & = & 132 \\ \end{array}$
Step 5. Solve the equation.	$\,\begin{array}{rll} {n}^{2}+n & = & 132 \\ \end{array}$
Bring all the terms to one side.	$\,\begin{array}{rll} {n}^{2}+n-132 & = & 0 \\ \end{array}$
Factor the trinomial.	$\begin{array}{rll} (n-11)(n+12) & = & 0 \\ \end{array}$
Use the zero product property.Solve the equations.	$\,\begin{array}{rllccccrll} \,n-11 & = & 0 & & & & & n+12 & = & 0 \\ n & = & 11 & & & & & n & = & -12 \\ \end{array}$
There are two values for $n$ that are solutions to this problem. So there are two sets of consecutive integers that will work.
$\begin{array}{rccccr} \mathrm{If the first integer is}\,n=11 & & & & & \mathrm{If the first integer is}\,n=-12 \\ \mathrm{then the next integer is}\,n+1 & & & & & \mathrm{then the next integer is}\,n+1 \\ 11+1 & & & & & -12+1 \\ 12 & & & & & -11 \\ \end{array}$
Step 6. Check the answer.
The consecutive integers are $11,12$ and $-11,-12$ . The product $11\cdot{}12=132$ and the product $-11(-12)=132$ . Both pairs of consecutive integers are solutions.
Step 7. Answer the question. The consecutive integers are $11,12$ and $-11,-12$ .

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.

17. Example 7.81

Example 7.81

A rectangular garden has an area $15$ square feet. The length of the garden is two feet more than the width. Find the length and width of the garden.

Solution

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

We will use this formula in the next example.

18. Example 7.82

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.

Solution

19. Section 7.6 Exercises

Section 7.6 Exercises

Use the Zero Product Property

In the following exercises, solve.

Solve Quadratic Equations by Factoring

In the following exercises, solve.

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve.

Mixed Practice

In the following exercises, solve.

Everyday Math

Writing Exercises

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

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

20. Practice 1

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

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

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

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

21. Practice 2

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

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

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

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

22. Practice 3

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

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

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

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

23. Practice 4

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

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

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

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

24. Practice 5

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

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

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

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

25. Practice 6

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

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

The product of two consecutive integers is $240$ . Find the integers.

The product of two consecutive integers is $420$ . Find the integers.

26. Practice 7

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.

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.

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.

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.

27. Practice 8

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

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

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

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

28. Practice 9

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

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

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

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

29. Practice 10

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

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

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

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

30. Practice 11

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

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

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

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.

31. Practice 12

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.

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

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

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

32. Practice 13

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

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

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 ${(s+4)}^{2}=196$ for s to find the length of a side of the patio.

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