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 635: 22


The solution set is {${-5,-4}$}.

Work Step by Step

$20x^{-2} + 9x^{-1} + 1 = 0$ Let $u=x^{-1}$ Rewrite the given equation as a quadratic equation in $u$ and solve for $u$. $20u^2 + 9u + 1 = 0$ Get the factors. $20u^2 + 9u + 1 = 0$: $(5u+1)(4u+1)=0$ Thus, $5u+1=0$ or $4u+1=0$ $u=-\frac{1}{5}$ or $u=-\frac{1}{4}$ Solve for $u$ $-\frac{1}{5}=x^{-1}$ and $-\frac{1}{4}=x^{-1}$ $-\frac{1}{5}=\frac{1}{x}$ and $-\frac{1}{4}=\frac{1}{x}$ Cross-multiply. $x=-5$ and $x=-4$ The solution set is {${-5,-4}$}.
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