Intermediate Algebra (6th Edition)

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

Chapter 8 - Review - Page 531: 30

Answer

$-5$

Work Step by Step

$x-(1/x)=-24/5$ $x<0$ $x-(1/x)=-24/5$ $x-(1/x)=-24/5$ $x*(x-(1/x))=-24/5*x$ $x^2-1=-24x/5$ $5*(x^2-1)=-24x/5*5$ $5x^2-5=-24x$ $5x^2-5+24x=-24x+24x$ $5x^2+24x-5=0$ $a=5$, $b=24$, $c=-5$ $x=(-b±\sqrt {b^2-4ac})/2a$ $x=(-24±\sqrt {24^2-4*5*(-5)})/2*5$ $x=(-24±\sqrt {576+100})/10$ $x=(-24±\sqrt {676})/10$ $x=(-24±26)/10$ $x=(-24±26)/10$ $x=(-24+26)/10$ $x=2/10$ $x=1/5$ This answer is not valid since we were looking for a negative number. $x=(-24±26)/10$ $x=(-24-26)/10$ $x=-50/10$ $x=-5$ $-5-(-1/5)$ $-25/5 - (-1/5)$ $-25/5 + 1/5$ $-24/5$
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