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McGraw Hill Integrated II, 2012

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McGraw Hill Integrated II, 2012 View details

5. Solving Quadratic Equations by Using the Quadratic Formula

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Exercise 34 Page 137

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

2

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 greatest common factor (GCF), if we find one.

12 - 12x = - 3x^2

LHS+3x^2=RHS+3x^2

12 - 12x + 3x^2 = 0

CommutativePropAdd

Commutative Property of Addition

3x^2 - 12x + 12 = 0

Factor out 3

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

.LHS /3.=.RHS /3.

x^2 - 4x + 4 = 0

Write as a sum

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

Distribute -1

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

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Factor out x & - 2

Factor out x

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

Factor out - 2

x(x - 2) - 2(x - 2) = 0

Factor out (x - 2)

(x - 2)(x - 2) = 0

ProdToPowTwoFac

a* a=a^2

(x-2)^2 = 0

Solving

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

(x-2)^2 = 0

Use the Zero Product Property

x - 2 = 0

LHS+2=RHS+2

x = 2

Checking Our Answer

Checking our answer

We can substitute our solution back into the given equation and simplify to check if our answers are correct.

12 - 12x = - 3x^2

x= 2

12 - 12( 2)? =- 3( 2)^2

▼

Simplify

Calculate power

12 - 12(2)? =- 3(4)

Multiply

12 - 24 ? = - 12

Subtract term

- 12 = - 12 ✓

We created a true statement. x=2 is a solution of the equation.

Subchapter links

Exercises p.137-139