Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.2 - Factoring the Difference of Two Squares - Problem Set 6.2 - Page 248: 53


The solution set is $\left\{-\frac{2}{5}, \frac{2}{5}\right\}$.

Work Step by Step

Add $-4$ to both sides to obtain: $25x^2-4=0 \\\\ (5x)^2-2^2=0$ RECALL: A difference of two squares can be factored using the formula: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares to obtain: $(5x-2)(5x+2)=0$ Equate each factor to 0, and then solve each equation to obtain: $5x-2=0 \text{ or } 5x+2 = 0 \\5x=2 \text{ or } 5x=-2 \\x=\frac{2}{5} \text{ or } x=-\frac{2}{5}$ The solution set is $\left\{-\frac{2}{5}, \frac{2}{5}\right\}$.
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