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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Using the Distributive Property
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Exercise 9 Page 39

Use the Zero Product Property to solve the equation.

0, - 10

Practice makes perfect

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

Solving

3k(k+10)=0 To do so, we will use the Zero Product Property.

3k(k+10)=0
ZeroProdProp

Use the Zero Product Property

lc3k=0 & (I) k+10=0 & (II)
SubEqn

(II): LHS-10=RHS-10

lc3k=0 & (I) k=-10 & (II)
DivEqn

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

lck=0 & (I) k=- 10 & (II)

We found that the solutions of the equation are 0 and - 10.

Checking

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

3k(k+10)=0
Substitute

k= 0

3( 0)( 0+10)? =0
ZeroPropMult

Zero Property of Multiplication

0(0+10)? =0
IdPropAdd

Identity Property of Addition

0(10)? =0
ZeroPropMult

Zero Property of Multiplication

0=0 ✓

Since substituting and solving resulted in a true statement, we know that k=0 is a solution of the equation. Let's now check k=- 10.

3k(k+10)=0
Substitute

k= -10

3( - 10)(( - 10)+10)? =0
AssociativePropAdd

Associative Property of Addition

3(- 10)(10-10)? =0
DiffZero

10-10=0

3(- 10)(0)? =0
ZeroPropMult

Zero Property of Multiplication

0=0 ✓

This is also a true statement, so we know that k=- 10 is a solution of the equation.

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Exercises 2 (Page 39)
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