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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

1. Solving Quadratic Equations by Factoring

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Exercise 59 Page 175

Make sure you rewrite the equation leaving all the terms on one side, and that you factor out the GCF if it exists.

x=1/2 and x=- 3

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start by writing all the terms on one side of the equals sign. We will also factor out a GCF, if we find one.

10x^2+25x=15

LHS-15=RHS-15

10x^2+25x-15=0

Factor out 5

5(2x^2+5x-3)=0

Write as a difference

5(2x^2+6x-x-3)=0

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Factor out 2x & - 1

Factor out 2x

5(2x(x+3)-x-3 )=0

Factor out - 1

5(2x(x+3)-(x+3) )=0

Factor out (x+3)

5(2x-1)(x+3)=0

Solving

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

5(2x-1)(x+3)=0

.LHS /5.=.RHS /5.

(2x-1)(x+3)=0

Use the Zero Product Property

lc2x-1=0 & (I) x+3=0 & (II)

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(I): Solve for x

(I): LHS+1=RHS+1

l2x=1 x+3=0

(I): .LHS /2.=.RHS /2.

lx= 12 x+3=0

(II): LHS-3=RHS-3

lx_1= 12 x_2=- 3

Subchapter links

Exercises p.174-177