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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

Before you get started, take this readiness quiz.

  1. Solve: 5 y 3 = 0 .
    If you missed this problem, review [link] .
  2. Solve: 10 a = 0 .
    If you missed this problem, review [link] .
  3. Combine like terms: 12 x 2 6 x + 4 x .
    If you missed this problem, review [link] .
  4. Factor n 3 9 n 2 22 n completely.
    If you missed this problem, review [link] .

We have already solved linear equations, equations of the form 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:

x 2 + 5 x + 6 = 0 3 y 2 + 4 y = 10 64 u 2 81 = 0 n ( n + 1 ) = 42

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 + b x + c = 0 , with a 0 .

Quadratic equation

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

a , b , and c are real numbers and a 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.

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 · b = 0 , then either a = 0 or b = 0 or both.

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

How to use the zero product property to solve a quadratic equation

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

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 next step is to solve each linear equation. This gives two solutions, x = −1 or x = 4. The last step is to check both answers by substituting the values for x into the original equation.
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Solve: ( x 3 ) ( x + 5 ) = 0 .

x = 3 , x = −5

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

y = 6 , y = −9

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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.

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

Solution

This image shows the steps for solving (5 n – 2)(6 n – 1) = 0. First, use the zero factor property to set each factor equal to 0, 5 n – 2 = 0 or 6 n – 1 = 0. Then, solve the equations, n = 2/5 or n = 1/6. Finally, check the answers by substituting the two solutions back into the original equation.
( 5 n 2 ) ( 6 n 1 ) = 0
Use the Zero Product Property to set
each factor to 0.
5 n 2 = 0 6 n 1 = 0
Solve the equations. n = 2 5 n = 1 6
Check your answers.
.

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

m = 2 3 , m = 1 2

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

p = 3 4 , p = 3 4

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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.

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

Solution

This image shows the steps for solving 3 p (10 p + 7) = 0. The first step is using the zero product property to set each factor equal to 0, 3p = 0 or 10 p + 7 = 0. The next step is solving both equations, p = 0 or p = negative 7/10. Finally, check the solutions by substituting the answers into the original equation.
3 p ( 10 p + 7 ) = 0
Use the Zero Product Property to set
each factor to 0.
3 p = 0 10 p + 7 = 0
Solve the equations. p = 0 10 p = −7
p = 7 10
Check your answers.
.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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