Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 111



Work Step by Step

$\displaystyle \frac{1}{x}+\frac{1}{x+2}=\frac{1}{3};$ First, exclude any x that yields 0 in a denominator $ x\neq 0,-2$ $\displaystyle \frac{1}{x}+\frac{1}{x+2}=\frac{1}{3} \qquad $ ... multiply with LCD = $3x(x+2)$ $ 3(x+2)+3x=x(x+2)\qquad $ ... write as $ ax^{2}+bx+c=0$ $3x+6+3x=x^{2}+2x $ $0=x^{2}-4x-6$ Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. $ x=\displaystyle \frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-6)}}{2(1)}$ $=\displaystyle \frac{4\pm\sqrt{16+24}}{2}=\frac{4\pm\sqrt{40}}{2}=\frac{4\pm\sqrt{4(10)}}{2}$ $=\displaystyle \frac{4\pm 2\sqrt{10}}{2}=\frac{2(2\pm\sqrt{10})}{2}$$=2\pm\sqrt{10}$ The solution set is $\{2+\sqrt{10},2-\sqrt{10}\}.$
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