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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 47 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.

- 1, 4

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.

2x^2-6x=8

LHS-8=RHS-8

2x^2-6x-8=0

▼

.LHS /2.=.RHS /2.

Factor out 2

2(x^2-3x-4)=0

.LHS /2.=.RHS /2.

x^2-3x-4=0

Write as a difference

x^2+x-4x-4=0

▼

Factor out x & (- 4)

Factor out x

x(x+1)-4x-4=0

Factor out (- 4)

x(x+1)-4(x+1)=0

Factor out (x+1)

(x+1)(x-4)=0

Solving

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

(x+1)(x-4)=0

Use the Zero Product Property

lcx+1=0 & (I) x-4=0 & (II)

(I): LHS-1=RHS-1

lx=- 1 x-4=0

(II): LHS+4=RHS+4

lx_1=- 1 x_2=4

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=- 1.

2x^2-6x=8

x= - 1

2( - 1)^2-6( - 1)? =8

▼

Evaluate left-hand side

Calculate power

2(1)-6(- 1)? =8

Identity Property of Multiplication

2-6(- 1)? =8

MultNegNegOnePar

- a(- b)=a* b

2+6? =8

Add terms

8=8 ✓

Substituting and simplifying created a true statement, so we know that x=- 1 is a solution of the equation. Let's move on to x=4.

2x^2-6x=8

x= 4

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

▼

Evaluate left-hand side

Calculate power

2(16)-6(4)? =8

Multiply

32-24? =8

Subtract term

8=8 ✓

Again, we created a true statement. x=4 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