Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 502: 48

Answer

$-12, 2$

Work Step by Step

$\frac{x}{x-5} + \frac{5}{x+5} = \frac{-1}{x^{2}-25}$ Using $(a^{2}-b^{2}) = (a+b)(a-b)$ formula, $x^{2}-25 = (x^{2}-5^{2}) = (x+5)(x-5)$ $\frac{x}{x-5} + \frac{5}{x+5} = \frac{-1}{(x+5)(x-5)}$ $x=5 $ or $x=-5$ makes the denominator zero. So, $\frac{x}{x-5} + \frac{5}{x+5} = \frac{-1}{(x+5)(x-5)}$;$ x\ne -5, $ or $x \ne 5$ Multiply both sides by $(x+5)(x-5)$ to clear fractions. $(x+5)(x-5)(\frac{x}{x-5} + \frac{5}{x+5}) =(x+5)(x-5)( \frac{-1}{(x+5)(x-5)})$ $(x+5)x+5(x-5) = -1 ; x\ne -5, $ or $x \ne 5$ $x^{2}+5x+5x-25=-1 $ $x^{2}+10x-25+1=0 $ $x^{2}+10x-24=0 $ By factoring, $(x+12)(x-2)=0$ $x=-12$ or $x=2$ Solutions are $-12,2$
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