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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 50 Page 230

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.

7, 1

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=8x-7

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Simplify

LHS-8x=RHS-8x

x^2-8x=- 7

LHS+7=RHS+7

x^2-8x+7=0

Write as a difference

x^2-x-7x+7=0

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Factor out x & (- 7)

Factor out x

x(x-1)-7x-7=0

Factor out (- 7)

x(x-1)-7(x-1)=0

Factor out (x-1)

(x-7)(x-1)=0

Solving

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

(x-7)(x-1)=0

Use the Zero Product Property

lcx-7=0 & (I) x-1=0 & (II)

(I): LHS+7=RHS+7

lx=7 x-1=0

(II): LHS+1=RHS+1

lx_1=7 x_2=1

Checking Our Answer

Checking our answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=7.

x^2=8x-7

x= 7

( 7)^2? =8( 7)-7

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Simplify

Calculate power

49? =8(7)-7

Multiply

49? =56-7

Subtract term

49=49 ✓

Substituting and simplifying created a true statement, so we know that x=7 is a solution of the equation. Let's move on to x=1.

x^2=8x-7

x= 1

1^2? =8( 1)-7

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Simplify

1^a=1

1? =8(1)-7

Identity Property of Multiplication

1? =8-7

Subtract term

1=1 ✓

Again, we created a true statement. x=1 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