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McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Using the Distributive Property
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Exercise 39 Page 40

Use the Zero Product Property to solve the equation.

0, 3

Practice makes perfect

We want to solve the given equation. At the end, we will check that our answer is correct.

Solving

3b(9b-27)=0 To do so, we will use the Zero Product Property.

3b(9b-27)=0
ZeroProdProp

Use the Zero Product Property

lc3b=0 & (I) 9b-27=0 & (II)
AddEqn

(II): LHS+27=RHS+27

lc3b=0 & (I) 9b=27 & (II)
DivEqn

(I): .LHS /3.=.RHS /3.

lcb=0 & (I) 9b=27 & (II)
DivEqn

(II): .LHS /9.=.RHS /9.

lcb=0 & (I) b=3 & (II)

We found that the solutions of the equation are 0 and 3.

Checking

To check our answers, we will substitute them for b in the given equation. Let's start with b=0.

3b(9b-27)=0
Substitute

b= 0

3( 0)(9(0)-27)? =0
ZeroPropMult

Zero Property of Multiplication

0(0-27)? =0
SubTerm

Subtract term

0(- 27)? =0
ZeroPropMult

Zero Property of Multiplication

0=0 ✓

Since substituting and solving resulted in a true statement, we know that b=0 is a solution of the equation. Let's now check b=3.

3b(9b-27)=0
Substitute

b= 3

3( 3)(9( 3)-27)? =0
Multiply

Multiply

9(27-27)? =0
DiffZero

27-27=0

9(0)? =0
ZeroPropMult

Zero Property of Multiplication

0=0 ✓

This is also a true statement, so we know that b=3 is a solution of the equation.

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Exercises 2 (Page 39)
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