## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 32

#### Answer

$x=\frac{2}{3}~~$ or $~~x=-\frac{3}{2}$

#### Work Step by Step

Let's consider a trinomial in this form: $~ax^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = a\times c$ Then the next step is to rewrite the trinomial as follows: $~x^2+bx+c = ax^2+rx+sx+c$ We can rewrite the given equation and use the GCF to factor it: $12x^2 +10x = 12$ $12x^2 +10x - 12 = 0$ $(2)~(6x^2 +5x - 6) = 0$ To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = 5~$ and $~r\times s = -36$. We can see that $(9)+(-4) = 5~$ and $(9)\times (-4) = -36$ We can solve the equation as follows: $12x^2 +10x = 12$ $12x^2 +10x - 12 = 0$ $(2)~(6x^2 +5x - 6) = 0$ $(2)~(6x^2 +9x - 4x - 6) = 0$ $(2)~[~(6x^2 +9x)+(- 4x - 6)~] = 0$ $(2)~[~(3x)(2x +3)+(-2)(2x +3)~] = 0$ $(2)~(3x-2)~(2x +3) = 0$ $3x-2 = 0~~$ or $~~2x+3 = 0$ $x=\frac{2}{3}~~$ or $~~x=-\frac{3}{2}$

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