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: 54


The solutions are $t=2+\sqrt{3}$ or $t=2-\sqrt{3}$.

Work Step by Step

$ t^{2}-4t=-1\qquad$..add $4$ to both sides to complete the square ($\displaystyle \frac{1}{2}(-4)=-2$, and $(-2)^{2}=4$.) $ t^{2}-4t+4=-1+4\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(t-2)^{2}=3$ 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-2=\pm\sqrt{3}\qquad$...add $2$ to each side. $ t-2+2=\pm\sqrt{3}+2\qquad$...simplify. $t=2\pm\sqrt{3}$ $t=2+\sqrt{3}$ or $t=2-\sqrt{3}$
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