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 8.2 - The Quadratic Formula - Exercise Set - Page 608: 45


$x=${$2 - \sqrt {10},2 + \sqrt {10}$}

Work Step by Step

Given: $\dfrac{1}{x}+\dfrac{1}{x+2}=\dfrac{1}{3}$ Re-write the given equation as: $x^2-4x-6=0$ Factorize the expression with the help of quadratic formula. Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ This implies that $x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(-6)}}{2(1)}$ or, $x=\dfrac{4 \pm \sqrt {40}}{2}$ or, $x=2 - \sqrt {10},2 + \sqrt {10}$ Hence, our solution set is: $x=${$2 - \sqrt {10},2 + \sqrt {10}$}
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