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: 11


$t=3+\sqrt{6}$ or $t=3-\sqrt{6}$

Work Step by Step

$ t^{2}+3=6t\qquad$...add $-6t$ to both sides. $ t^{2}-6t+3=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-6,\ c=3$ $ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-6,\ a$ for $1$ and $c$ for $3$. $ t=\displaystyle \frac{-(-6)\pm\sqrt{(-6)^{2}-4\cdot(3)\cdot 1}}{2\cdot 1}\qquad$... simplify. $t=\displaystyle \frac{6\pm\sqrt{36-12}}{2}$ $t=\displaystyle \frac{6\pm\sqrt{24}}{2}$ $t=\displaystyle \frac{6\pm\sqrt{4\cdot 6}}{2}$ $t=\displaystyle \frac{6\pm 2\sqrt{6}}{2}$ $ t=3\pm\sqrt{6}\qquad$... the symbol $\pm$ indicates two solutions. $t=3+\sqrt{6}$ or $t=3-\sqrt{6}$
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