# Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 501: 32

The solutions are $-1-\sqrt{5}$ and $-1+\sqrt{5}$.

#### Work Step by Step

Factor the denominator of the expression on the right side of the equation to obtain: $$\dfrac{5}{x-3} + \dfrac{x}{x+3} = \dfrac{19}{(x-3)(x+3)}$$ Multiply the LCD $(x-3)(x+3)$ to both sides of the equation to get rid of the denominators: \begin{array}{ccc} &(x-3)(x+3)\left[\dfrac{5}{x-3}+\dfrac{x}{x+3}\right] &= &\dfrac{19}{(x-3)(x+3)} \cdot (x-3)(x+3) \\&(x-3)(x+3) \cdot \dfrac{5}{x-3}+(x-3)(x+3) \cdot \dfrac{x}{x+3} &= &\dfrac{19}{(x-3)(x+3)} \cdot (x-3)(x+3) \end{array} Simplify by cancelling common factors to obtain: \begin{array}{ccc} \require{cancel} \\&\cancel{(x-3)}(x+3) \cdot \dfrac{5}{\cancel{x-3}}+(x-3)\cancel{(x+3)} \cdot \dfrac{x}{\cancel{x+3}} &= &-\dfrac{19}{\cancel{(x-3)(x+3)}} \cdot \cancel{(x-3)(x+3)} \\&(x+3)(5)+(x-3)(x) &= &19 \\&5(x) + 5(3) + x(x)-x(3) &= &19 \\&5x+15+x^2-3x &= &19 \\&x^2+2x+15 &= &19 \end{array} Put all terms on the left side. Note that when a term is transferred to the other side of an equation, it changes it sign to its opposite. \begin{array}{ccc} \\&x^2+2x+15 &= &19 \\&x^2+2x+15-19 &= &0 \\&x^2+2x-4 &= &0 \end{array} Solve the equation using the quadratic formula with $a=1, b=2, c=-4$ to obtain: $x=\dfrac{-2\pm \sqrt{2^2-4(1)(-4)}}{2(1)} \\x=\dfrac{-2\pm\sqrt{4+16}}{2} \\x=\dfrac{-2\pm \sqrt{20}}{2} \\x=\dfrac{-2\pm \sqrt{4(5)}}{2} \\x=\dfrac{-2\pm 2\sqrt{5}}{2} \\x=\dfrac{-2}{2} \pm \dfrac{2\sqrt{5}}{2} \\x=-1\pm \sqrt5$ Therefore, the solutions are $-1-\sqrt{5}$ and $-1+\sqrt{5}$.

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