Elementary Algebra

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



The solution set is $\left\{-4, 4\right\}$.

Work Step by Step

Add $-16$ to both sides to obtain: $x^2-16=0 \\x^2-4^2=0$ 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: $(x-4)(x+4)=0$ Equate each factor to 0, and then solve each equation to obtain: $x-4=0 \text{ or } x+4 = 0 \\x=4 \text{ or } x=-4$ The solution set is $\left\{-4, 4\right\}$.
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