Algebra 2 (1st Edition)

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

Chapter 4 Quadratic Functions and Factoring - 4.7 Complete the Square - 4.7 Exercises - Skill Practice - Page 289: 55


The solutions are $\displaystyle \frac{1}{6}+i\frac{\sqrt{71}}{6}$ and $ \displaystyle \frac{1}{6}-i\frac{\sqrt{71}}{6}$

Work Step by Step

$ 3x^{2}+x=2x-6\qquad$ ...add $-2x$ to each side $ 3x^{2}-x=-6\qquad$ ...divide the entire expression with $3$. $ x^{2}-\displaystyle \frac{1}{3}x=-2\qquad$ ...square half the coefficient of $q$. $(\displaystyle \frac{\frac{-1}{3}}{2})^{2}=(\frac{1}{6})^{2}\qquad$ ...complete the square by adding$ \displaystyle \frac{1}{36}$ to each side of the expression $ x^{2}-\displaystyle \frac{1}{3}x+\frac{1}{36}=-2+\frac{1}{36}\qquad$ ...Write $x^{2}-\displaystyle \frac{1}{3}x+\frac{1}{36}$ as a binomial squared. $(x-\displaystyle \frac{1}{6})^{2}=-\frac{71}{36}\qquad$ ...take square roots of each side. $ x-\displaystyle \frac{1}{6}=\pm\sqrt{-\frac{71}{36}}\qquad$ ...simplify $\displaystyle \sqrt{-\frac{71}{36}}=i\sqrt{\frac{71}{36}}=i\frac{\sqrt{71}}{\sqrt{36}}=i\frac{\sqrt{71}}{6}$ $ x-\displaystyle \frac{1}{6}=\pm i\frac{\sqrt{71}}{6}\qquad$ ...add $\displaystyle \frac{1}{6}$ to each side. $x=\displaystyle \frac{1}{6}\pm i\frac{\sqrt{71}}{6}$
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.