Elementary Algebra

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

Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set: 13

Answer

{$-3\sqrt 5,3\sqrt 5$}

Work Step by Step

Using Property 10.1, which says that for any non-negative real number $a$, $x^{2}=a$ can be written as $x=\pm\sqrt a$, we obtain: Step 1: $y^{2}=45$ Step 2: $y=\pm \sqrt {45}$ Step 3: $y=\pm \sqrt {9\times5}$ Step 4: $y=\pm (\sqrt 9\times\sqrt 5)$ Step 5: $y=\pm (3\times\sqrt 5)$ Step 6: $y=\pm (3\sqrt 5)$ Step 7: $y=3\sqrt 5$ or $y=-3\sqrt 5$ The solution set is {$-3\sqrt 5,3\sqrt 5$}.
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