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: 61


The solution set is $\left\{-\frac{1}{3}, \frac{1}{3}\right\}$.

Work Step by Step

Add $45x^2-5$ to both sides to find: $0=45x^2-5$ Divide by $5$ on both sides to find: $0=9x^2-1 \\0=(3x)^2-1^{2}$ 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 find: $(3x-1)(3x+1)=0$ Equate each factor to 0 then solve each equation to find: $3x-1=0 \text{ or } 3x+1 = 0\\\ 3x=1 \text{ or } 3x=-1 \\x=\frac{1}{3} \text{ or }x=-\frac{1}{3}$ The solution set is $\left\{-\frac{1}{3}, \frac{1}{3}\right\}$.
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