Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.6 - Other Types of Equations - 1.6 Exercises - Page 128: 23



Work Step by Step

Let's find the solutions as follows. $\begin{align}\frac{1}{x^2}+\frac{8}{x}+15&=0\\\frac{1+8x+15x^2}{x^2}&=0\\15x^2+8x+1&=0\\15x^2+3x+5x+1&=0\\3x(5x+1)+1(5x+1)&=0\\(5x+1)(3x+1)&=0\\x&=-\frac{1}{5} \text{ or } x=-\frac{1}{3}\end{align}$ Now to check substitute $x=-\frac{1}{5}$ and $x=-\frac{1}{3}$ in original equation. $\frac{1}{\left(\frac{-1}{5}\right)^2}+\frac{8}{\frac{-1}{5}}+15=25-40+15=0$ $\frac{1}{\left(\frac{-1}{3}\right)^2}+\frac{8}{\frac{-1}{3}}+15=9-24+15=0$ Hence the solutions are $x=-\frac{1}{3},x=-\frac{1}{5}$.
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