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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set: 45

Answer

$z=\left\{ 0,\dfrac{4}{3} \right\}$

Work Step by Step

Using the properties of equality, the given expression, $ 15z^2+7=20z+7 ,$ is equivalent to \begin{array}{l}\require{cancel} 15z^2-20z+7-7=0 \\\\ 15z^2-20z=0 .\end{array} Factoring the above equation, $ 15z^2-20z=0 ,$ results to \begin{array}{l}\require{cancel} 5z(3z-4) .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ 5z(3z-4) ,$ are \begin{array}{l}\require{cancel} 5z=0 \\\\ z=\dfrac{0}{5} \\\\ z=0 ,\\\\\text{OR}\\\\ 3z-4=0 \\\\ 3z=4 \\\\ z=\dfrac{4}{3} .\end{array} Hence, $ z=\left\{ 0,\dfrac{4}{3} \right\} .$
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