Intermediate Algebra (11th edition)

Solving a Quadratic Equation with a Missing Constant Term

Solve

This quadratic equation has a missing term. Comparing it with the standard form

shows that The zero-factor property can still be used.

Factor.

or Zero-factor property

x= 0 or x= 5 Solve each equation.

4 x= 0 x - 5 = 0

4 x 1 x- 52 = 0

4 x^2 - 20 x= 0

ax^2 +bx+ c= 0 c= 0.

4 x^2 - 20 x=0.

EXAMPLE 3

346 CHAPTER 6 Factoring

NOW TRY
EXERCISE 3
Solve 3x^2 + 12 x=0.

NOW TRY ANSWERS

3. 5 0, - 46 4. 5 - 5, 5 6

Set each variable

factor equal to 0.

CHECK

Let. Let.

✓ True ✓ True

The solution set is 5 0, 5 6. NOW TRY

0 - 0 = 0 100 - 100 = 0

41022 - 20102  0 x= 0 41522 - 20152  0 x= 5

4 x^2 - 20 x= 0 4 x^2 - 20 x= 0

CAUTION Remember to include 0 as a solution in Example 3.

Solving a Quadratic Equation with a Missing Linear Term

Solve

Factor out 3.

Factor

or Zero-factor property

x=- 6 or x= 6

x+ 6 = 0 x - 6 = 0

31 x+ 621 x- 62 = 0 x^2 - 36.

31 x^2 - 362 = 0

3 x^2 - 108 = 0

3 x^2 - 108 = 0.

NOW TRY EXAMPLE 4

EXERCISE 4
Solve 4x^2 - 100 =0.

The factor 3

does notlead to

a solution.

Check that the solution set is 5 - 6, 6 6. NOW TRY

CAUTION The factor 3 in Example 4is not a variablefactor, so it does notlead

to a solution of the equation. In Example 3,however, the factor xis a variable factor

and leads to the solution 0.

Solving an Equation That Requires Rewriting

Solve

Multiply on each side.

Add 2x. Subtract 6.

Factor.

or Zero-factor property

x=- or x= 1 Solve each equation.

7

2

2 x+ 7 = 0 x - 1 = 0

12 x+ 721 x- 12 = 0

2 x^2 + 5 x- 7 = 0

2 x^2 + 3 x+ 1 = 2 - 2 x+ 6

12 x+ 121 x+ 12 = 211 - x 2 + 6

12 x+ 121 x+ 12 = 211 - x 2 + 6.

EXAMPLE 5

Write in standard form.