Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.4 - Solving Quadratic Equations - Which Method? - Problem Set 10.4 - Page 457: 4



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: $(3n-1)^{2}=25$ Step 2: $3n-1=\pm \sqrt {25}$ Step 3: $3n-1=\pm 5$ Step 4: $3n-1=5$ or $3n-1=-5$ Step 5: $3n=5+1$ or $3n=-5+1$ Step 6: $3n=6$ or $3n=-4$ Step 7: $n=\frac{6}{3}=2$ or $n=-\frac{4}{3}$ The solution set is {$-\frac{4}{3},2$}.
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.