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: 112


$\displaystyle \{\frac{5+\sqrt{73}}{2},\frac{5-\sqrt{73}}{2}\}.$

Work Step by Step

First, exclude any x that yields 0 in a denominator $ x\neq 0,-3$ $\displaystyle \frac{1}{x}+\frac{1}{x+3}=\frac{1}{4} \qquad $ ... multiply with LCD = $4x(x+3)$ $ 4(x+3)+4x=x(x+3)\qquad $ ... write as $ ax^{2}+bx+c=0$ $4x+12+4x=x^{2}+3x $ $0=x^{2}-5x-12$ Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $ x=\displaystyle \frac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(-12)}}{2(1)}$ $=\displaystyle \frac{5\pm\sqrt{25+48}}{2}=\frac{5\pm\sqrt{73}}{2}$ The solution set is $\displaystyle \{\frac{5+\sqrt{73}}{2},\frac{5-\sqrt{73}}{2}\}.$
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