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: 16


$t=-1+\displaystyle \frac{\sqrt{6}}{2}$ or $t=-1-\displaystyle \frac{\sqrt{6}}{2}$

Work Step by Step

$ 2t(t+2)=1\qquad$... use the distributive property: $a(b+c)=ab+ac$. $2t\cdot t+2t\cdot 2=1$ $ 2t^{2}+4t=1\qquad$...add $-1$ to both sides. $ 2t^{2}+4t-1=0\qquad$... solve with the Quadractic formula. $a=2,\ b=4,\ c=-1$ $ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $4,\ a$ for $2$ and $c$ for $-1$. $ t=\displaystyle \frac{-4\pm\sqrt{(4)^{2}-4\cdot(-1)\cdot 2}}{2\cdot 2}\qquad$... simplify. $t=\displaystyle \frac{-4\pm\sqrt{16+8}}{4}$ $t=\displaystyle \frac{-4\pm\sqrt{24}}{4}$ $t=\displaystyle \frac{-4\pm\sqrt{4\cdot 6}}{4}$ $t=\displaystyle \frac{-4\pm 2\sqrt{6}}{4}$ $ t=-1\displaystyle \pm\frac{\sqrt{6}}{2}\qquad$... the symbol $\pm$ indicates two solutions. $t=-1+\displaystyle \frac{\sqrt{6}}{2}$ or $t=-1-\displaystyle \frac{\sqrt{6}}{2}$
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