Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.5 Solve Quadratic Equations by Finding Square Roots - 4.5 Exercises - Skill Practice - Page 270: 33


The solutions are $-2+\displaystyle \frac{\sqrt{26}}{2}$ and $-2-\displaystyle \frac{\sqrt{26}}{2}$.

Work Step by Step

$ 2(x+2)^{2}-5=8\qquad$ ...add $5$ to each side. $ 2(x+2)^{2}-5+5=8+5\qquad$ ...simplify. $ 2(x+2)^{2}=13\qquad$ ...divide each side with $2$. $(x+2)^{2}=\displaystyle \frac{13}{2}\qquad$ ...take square roots of each side. $\sqrt{(x+2)^{2}}=\sqrt{\frac{13}{2}}\qquad$ ...simplify. ...When solving an equation of the form $x^{2}=s$ where $s>0$, we find both the positive and negative solutions. $ x+2=\pm\sqrt{\frac{13}{2}}\qquad$ ...apply the Quotient Property:$\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ $ x+2=\displaystyle \pm\frac{\sqrt{13}}{\sqrt{2}}\qquad$ ...rationalize the denominator by multiplying both the numerator and the denominator with $\sqrt{2}$ $ x+2=\displaystyle \pm\frac{\sqrt{13}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\qquad$ ...use the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $ x+2=\displaystyle \pm\frac{\sqrt{26}}{\sqrt{4}}\qquad$ ...evaluate $\sqrt{4} $ ($\sqrt{4}=2$) $ x+2=\displaystyle \pm\frac{\sqrt{26}}{2}\qquad$ ...add $-2$ to each side. $ x+2-2=\displaystyle \pm\frac{\sqrt{26}}{2}-2\qquad$ ...simplify. $x=-2\displaystyle \pm\frac{\sqrt{26}}{2}$
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.