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 288: 24


The solutions are $-3+2\sqrt{3}$ and $-3-2\sqrt{3}$.

Work Step by Step

$ x^{2}+6x-3=0\qquad$ ... write left side in the form $x^{2}+bx$ (add $3$ to each side). $ x^{2}+6x-3+3=0+3\qquad$ ...simplify. $ x^{2}+6x=3\qquad$ ...square half the coefficient of $x$. $(\displaystyle \frac{6}{2})^{2}=3^{2}=9\qquad$ ...add $9$ to each side of the expression $ x^{2}+6x+9=3+9\qquad$ ...simplify. $ x^{2}+6x+9=12\qquad$ ... write left side as a binomial squared. $(x+3)^{2}=12\qquad$ ...take square roots of each side. $ x+3=\pm\sqrt{12}\qquad$ ...rewrite $\sqrt{12}$ as $\sqrt{4\cdot 3.}$ $ x+3=\pm\sqrt{4\cdot 3.}\qquad$ ...evaluate $\sqrt{4}$. $ x+3=\pm 2\sqrt{3}\qquad$ ...add $-3$ to each side. $ x+3-3=\pm 2\sqrt{3}-3\qquad$ ...simplify. $x=-3\pm 2\sqrt{3}$
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