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: 15


$t=-1+\displaystyle \frac{2\sqrt{3}}{3}$ or $t=-1-\displaystyle \frac{2\sqrt{3}}{3}$

Work Step by Step

$ 3t(t+2)=1\qquad$... use the distributive property: $a(b+c)=ab+ac$. $3t\cdot t+3t\cdot 2=1$ $ 3t^{2}+6t=1\qquad$...add $-1$ to both sides. $ 3t^{2}+6t-1=0\qquad$... solve with the Quadractic formula. $a=3,\ b=6,\ c=-1$ $ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $6,\ a$ for $3$ and $c$ for $-1$. $ t=\displaystyle \frac{-6\pm\sqrt{(6)^{2}-4\cdot(-1)\cdot 3}}{2\cdot 3}\qquad$... simplify. $t=\displaystyle \frac{-6\pm\sqrt{36+12}}{6}$ $t=\displaystyle \frac{-6\pm\sqrt{48}}{6}$ $t=\displaystyle \frac{-6\pm\sqrt{4\cdot 4\cdot 3}}{6}$ $t=\displaystyle \frac{-6\pm 4\sqrt{3}}{6}$ $ t=-1\displaystyle \pm\frac{2\sqrt{3}}{3}\qquad$... the symbol $\pm$ indicates two solutions. $t=-1+\displaystyle \frac{2\sqrt{3}}{3}$ or $t=-1-\displaystyle \frac{2\sqrt{3}}{3}$
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