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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

5. Quadratic Equations

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Exercise 9 Page 229

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

- 4, - 2

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start by factoring the left-hand side of the equation.

x^2 +6x+8=0

Write as a sum

x^2 +2x+4x+8=0

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Factor out x & 2

Factor out x

x(x+2)+4x+8=0

Factor out 4

x(x+2)+4(x+2)=0

Factor out (x+2)

(x+4)(x+2)=0

Solving

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

(x+4)(x+2)=0

Use the Zero Product Property

lcx+4=0 & (I) x+2=0 & (II)

(I): LHS-4=RHS-4

lx=- 4 x+2=0

(II): LHS-2=RHS-2

lx_1=- 4 x_2=- 2

Checking

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=- 4.

x^2+6x+8=0

x= - 2

( - 4)^2+6( - 4)+8? =0

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Evaluate left-hand side

Calculate power

16+6(- 4)+8? =0

a(- b)=- a * b

16-24+8? =0

Add and subtract terms

0=0 ✓

Substituting and simplifying created a true statement, so we know that x=- 4 is a solution of the equation. Let's move on to x=- 2.

x^2+6x+8=0

x= - 2

( - 2)^2+6( - 2)+8? =0

▼

Evaluate left-hand side

Calculate power

4+6(- 2)+8? =0

a(- b)=- a * b

4-12+8? =0

Add and subtract terms

0=0 ✓

Again, we created a true statement. x=- 2 is indeed a solution of the equation.

Subchapter links

Lesson Check p.229

Practice and Problem-Solving Exercises p.229-231

Standardized Test Prep p.231

Mixed Review p.231