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: 53


The solutions are $t=5+\sqrt{2}$ or $t=5-\sqrt{2}$

Work Step by Step

$ t^{2}-10t=-23\qquad$..add $25$ to both sides to complete the square ($\displaystyle \frac{1}{2}(-10)=-5$, and $(-5)^{2}=25$.) $ t^{2}-10t+25=-23+25\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(t-5)^{2}=2$ 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-5=\pm\sqrt{2}\qquad$...add $5$ to each side. $ t-5+5=\pm\sqrt{2}+5\qquad$...simplify. $t=5\pm\sqrt{2}$ $t=5+\sqrt{2}$ or $t=5-\sqrt{2}$
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