Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 82


Solution set = $\displaystyle \{\frac{3-\sqrt{6}}{3},\frac{3+\sqrt{6}}{3}\}.$

Work Step by Step

Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ Add $-6x+1$ to both sides, so we have $ax^{2}+bx+c=0.$ $ 3x^{2}-6x+1=0\quad\rightarrow \left\{\begin{array}{l} a=3\\ b=-6\\ c=1 \end{array}\right.$ $x=\displaystyle \frac{6\pm\sqrt{(-6)^{2}-4(3)(1)}}{2(3)}=\frac{6\pm\sqrt{36-12}}{6}=\frac{6\pm\sqrt{24}}{6}$ $=\displaystyle \frac{6\pm\sqrt{4\cdot 6}}{6}=\frac{6\pm 2\sqrt{6}}{6}=\frac{2(3\pm\sqrt{6})}{6}$ $=\displaystyle \frac{3\pm\sqrt{6}}{3}$ Solution set = $\displaystyle \{\frac{3-\sqrt{6}}{3},\frac{3+\sqrt{6}}{3}\}.$
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