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 - Page 351: 11


$x=\left\{ \dfrac{1}{7},\dfrac{3}{10} \right\}$

Work Step by Step

Dividing both sides by $ 4 ,$ the given equation, $ 4(7x-1)(10x-3)=0 ,$ is equivalent to \begin{array}{l} (7x-1)(10x-3)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (7x-1)(10x-3)=0 ,$ is \begin{array}{l} 7x-1=0 \\\\ 7x=1 \\\\ x=\dfrac{1}{7} ,\\\\\text{OR}\\\\ 10x-3=0 \\\\ 10x=3 \\\\ x=\dfrac{3}{10} .\end{array} Hence, $ x=\left\{ \dfrac{1}{7},\dfrac{3}{10} \right\} .$
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