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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 10 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.

6, 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.

x^2+18=9x

LHS-9x=RHS-9x

x^2-9x+18=0

Write as a difference

x^2-3x-6x+18=0

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Factor out x & - 6

Factor out x

x(x-3)-6x+18=0

Factor out (- 6)

x(x-3)-6(x-3)=0

Factor out (x-3)

(x-6)(x-3)=0

Solving

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

(x-6)(x-3)=0

Use the Zero Product Property

lcx-6=0 & (I) x-3=0 & (II)

(I): LHS+6=RHS+6

lx=6 x-3=0

(II): LHS+3=RHS+3

lx_1=6 x_2=3

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=6.

x^2+18=9x

x= 6

6^2+18? =9( 6)

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Simplify

Calculate power

36+18? =9(6)

Add terms

54? =9(6)

Multiply

54=54 ✓

Substituting and simplifying created a true statement, so we know that x=6 is a solution of the equation. Let's move on to x=3.

x^2+18=9x

x= 3

3^2+18? =9( 3)

▼

Simplify

Calculate power

9+18? =9(3)

Add terms

27? =9(3)

Multiply

27=27 ✓

Again, we created a true statement. x=3 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