Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 636: 32



Work Step by Step

Since, $u^2+6u-27=0$ Factorize the expression as follows: $(u+9)(u-3)=0$ or, $u=${$-9,3$} Replace $u$ with $(x-\dfrac{10}{x})$, we have $(x-\dfrac{10}{x})=-9$ or, $x^2+9x-10=0$ or,$(x+10)(x-1)=0$ or, $x=${$-10,1$} and $(x-\dfrac{10}{x})=3$ or, $x^2-3x-10=0$ or,$(x+2)(x-5)=0$ or, $x=${$-2,5$} Hence, our solution set is $x=${$-10,-2,1,5$}
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