## Precalculus (6th Edition) Blitzer

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