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 - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 706: 58


The solutions are $t=-3+\sqrt{14}$ or $t=-3-\sqrt{14}$.

Work Step by Step

$ t^{2}+6t-5=0\qquad$..add $5$ to both sides so we can complete the square on the left side. $ t^{2}+6t=5\qquad$..add $9$ to both sides to complete the square ($\displaystyle \frac{1}{2}(6)=3$, and $(3)^{2}=9$.) $ t^{2}+6t+9=5+9\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(t+3)^{2}=14$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $ t+3=\pm\sqrt{14}\qquad$...add $-3$ to each side. $ t+3-3=\pm\sqrt{14}-3\qquad$...simplify. $t=-3\pm\sqrt{14}$ $t=-3+\sqrt{14}$ or $t=-3-\sqrt{14}$
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