McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Solving Quadratic Equations by Using the Quadratic Formula
Continue to next subchapter

Exercise 30 Page 137

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.

2, 6

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.
3x^2 - 24x = - 36
3x^2- 24x+36=0
3(x^2 - 8x + 12) = 0
x^2 - 8x + 12 = 0
x^2 - (2x + 6x) + 12 = 0
x^2 - 2x - 6x + 12 = 0
â–Ľ
Factor out x & -6
x(x - 2) - 6x + 12
x(x - 2) - 6(x - 2)
(x-2)(x - 6)

Solving

To solve this equation, we will apply the Zero Product Property.
(x-2)(x-6)=0
lcx-2=0 & (I) x-6=0 & (II)
lx=2 x-6=0
lx=2 x=6

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=2.
3x^2 - 24x = - 36
3( 2)^2- 24( 2)? =- 36
â–Ľ
Simplify
3(4)- 24(2)? =- 36
12 - 48 ? =- 36
- 36=-36 âś“
Substituting and simplifying created a true statement, so we know that x=2 is a solution of the equation. Let's move on to x=6.
3x^2 - 24x = - 36
3( 6)^2-24( 6)? =- 36
â–Ľ
Simplify
3(36) - 24(6)? =- 36
108 - 144 ? =- 36
- 36=-36 âś“
Again, we created a true statement. x=6 is indeed a solution of the equation.