## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set - Page 351: 12

#### Answer

$x=\left\{ -\dfrac{9}{2},\dfrac{3}{4} \right\}$

#### Work Step by Step

Dividing both sides by $6 ,$ the given equation, $6(4x-3)(2x+9)=0 ,$ is equivalent to \begin{array}{l} (4x-3)(2x+9)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $(4x-3)(2x+9)=0 ,$ is \begin{array}{l} 4x-3=0 \\\\ 4x=3 \\\\ x=\dfrac{3}{4} ,\\\\\text{OR}\\\\ 2x+9=0 \\\\ 2x=-9 \\\\ x=-\dfrac{9}{2} .\end{array} Hence, $x=\left\{ -\dfrac{9}{2},\dfrac{3}{4} \right\} .$

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