## 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 352: 46

#### Answer

$z=\left\{ 0,\dfrac{3}{2} \right\}$

#### Work Step by Step

Using the properties of equality, the given expression, $14z^2-3=21z-3 ,$ is equivalent to \begin{array}{l}\require{cancel} 14z^2-21z-3+3=0 \\\\ 14z^2-21z=0 .\end{array} Factoring the above equation, $14z^2-21z=0 ,$ results to \begin{array}{l}\require{cancel} 7z(2z-3)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $7z(2z-3)=0 ,$ are \begin{array}{l}\require{cancel} 7z=0 \\\\ z=\dfrac{0}{7} \\\\ z=0 ,\\\\\text{OR}\\\\ 2z-3=0 \\\\ 2z=3 \\\\ z=\dfrac{3}{2} .\end{array} Hence, $z=\left\{ 0,\dfrac{3}{2} \right\} .$

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