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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 44 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.

3.5, - 4

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.

2x^2+x-28=0

Write as a difference

2x^2+8x-7x-28=0

▼

Factor out 2x & - 7

Factor out 2x

2x(x+4)-7x-28=0

Factor out (- 7)

2x(x+4)-7(x+4)=0

Factor out (x+4)

(2x-7)(x+4)=0

Solving

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

(2x-7)(x+4)=0

Use the Zero Product Property

l2x-7=0 x+4=0

(I): LHS+7=RHS+7

l2x=7 x+4=0

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

lx=3.5 x+4=0

(II): LHS-4=RHS-4

lx_1=3.5 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=3.5.

2x^2+x-28=0

x= 3.5

2( 3.5)^2+ 3.5-28? =0

▼

Evaluate left-hand side

Calculate power

2(12.25)+3.5-28? =0

Multiply

24.5+3.5-28? =0

Add and subtract terms

0=0 ✓

Substituting and simplifying created a true statement, so we know that x=3.5 is a solution of the equation. Let's move on to x=- 4.

2x^2+x-28=0

x= - 4

2( - 4)^2+( - 4)-28? =0

▼

Evaluate left-hand side

Calculate power

2(16)+(- 4)-28? =0

Multiply

32+(- 4)-28? =0

a+(- b)=a-b

32-4-28? =0

Subtract terms

0=0 ✓

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