Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 10


$u=1+\sqrt{3}$ or $u=1-\sqrt{3}$

Work Step by Step

$ u^{2}-2u-2=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-2,\ c=-2$ $ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-2,\ a$ for $1$ and $c$ for $-2$. $ u=\displaystyle \frac{-(-2)\pm\sqrt{(-2)^{2}-4\cdot(-2)\cdot 1}}{2\cdot 1}\qquad$... simplify. $u=\displaystyle \frac{2\pm\sqrt{4+8}}{2}$ $u=\displaystyle \frac{2\pm\sqrt{12}}{2}$ $u=\displaystyle \frac{2\pm\sqrt{4\cdot 3}}{2}$ $u=\displaystyle \frac{2\pm 2\sqrt{3}}{2}$ $ u=1\pm\sqrt{3}\qquad$... the symbol $\pm$ indicates two solutions. $u=1+\sqrt{3}$ or $u=1-\sqrt{3}$
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