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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

Mid-Chapter Quiz

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Exercise 12 Page 188

Rewrite the x-term as a sum of two x-terms. Then apply the Zero Product Property.

x=- 3 and x=5

We want to solve the given equation by factoring.

Factoring

Let's start by rewriting - 2x as the sum of - 5x and 3x.

x^2-2x-15=0

Write as a sum

x^2-5x+3x-15=0

Factor out x

x(x-5)+3x-15=0

Factor out 3

x(x-5)+3(x-5)=0

Factor out (x-5)

(x+3)(x-5)=0

Solving

To solve the equation, we will apply the Zero Product Property.

(x+3)(x-5)=0

Use the Zero Product Property

lcx+3=0 & (I) x-5=0 & (II)

(I): LHS-3=RHS-3

lx=- 3 x-5=0

(II): LHS+5=RHS+5

lx_1=- 3 x_2=5

We found that the two solutions to our equation are x=- 3 and x=5.

Checking Our Answer

Checking Our Answer

We can check our solutions by substituting them into the given equation. If the substitution produces a true statement, our solution is correct. Let's start with x=- 3.

x^2-2x-15=0

x= - 3

( -3)^2-2( -3)-15? =0

▼

Simplify

Calculate power

9-2(-3)-15? =0

MultNegNegOnePar

- a(- b)=a* b

9+6-15? =0

Add and subtract terms

0=0 ✓

Since our substitution produced a true statement, x=- 3 is a correct solution. Let's move on to x=5.

x^2-2x-15=0

x= 5

( 5)^2-2( 5)-15? =0

▼

Simplify

Calculate power

25-2(5)-15? =0

(- a)b = - ab

25-10-15? =0

Subtract terms

0=0 ✓

Our substitution produced a true statement again. Therefore, x=5 is also a correction solution.