Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.1 - Quadratic Equations - Problem Set 10.1 - Page 442: 32

Answer

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

Work Step by Step

Using Property 10.1, which states that for any non-negative real number $a$, $x^{2}=a$ can be written as $x=\pm\sqrt a$, we obtain: Step 1: $4x^{2}-108=0$ Step 2: $4x^{2}=108$ Step 3: $x^{2}=\frac{108}{4}$ Step 4: $x^{2}=27$ Step 5: $x=\pm \sqrt {27}$ Step 6: $x=\pm \sqrt {9\times 3}$ Step 7: $x=\pm (\sqrt {9}\times \sqrt 3)$ Step 8: $x=\pm (3\times \sqrt 3)$ Step 9: $x=\pm 3\sqrt 3$ The solution set is {$-3\sqrt 3, 3\sqrt 3$}.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.