## Precalculus (6th Edition) Blitzer

The solution set of equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$ is $\left\{ \left. 4 \right\} \right.$.
Consider the provided equation, $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$ Multiply both sides by ${{x}^{2}}+6x+8$ \begin{align} & \frac{2x}{{{x}^{2}}+6x+8}\cdot \left( {{x}^{2}}+6x+8 \right)+\frac{2}{x+2}\cdot \left( {{x}^{2}}+6x+8 \right)=\frac{x}{x+4}\cdot \left( {{x}^{2}}+6x+8 \right) \\ & 2x+\frac{2}{x+2}\cdot \left( {{x}^{2}}+6x+8 \right)=\frac{x}{x+4}\cdot \left( {{x}^{2}}+6x+8 \right) \\ \end{align} On factorization of ${{x}^{2}}+6x+8$ it becomes \begin{align} & {{x}^{2}}+6x+8={{x}^{2}}+4x+2x+8 \\ & =x\left( x+4 \right)+2\left( x+4 \right) \\ & =\left( x+4 \right)\left( x+2 \right) \end{align} Now replace ${{x}^{2}}+6x+8$ by $\left( x+4 \right)\left( x+2 \right)$ \begin{align} & 2x+\frac{2}{x+2}\cdot \left( x+4 \right)\left( x+2 \right)=\frac{x}{x+4}\cdot \left( x+4 \right)\left( x+2 \right) \\ & 2x+2\cdot \left( x+4 \right)=x\cdot \left( x+2 \right) \\ & 2x+2x+8={{x}^{2}}+2x \\ & 4x+8={{x}^{2}}+2x \end{align} On further simplification \begin{align} & 8={{x}^{2}}+2x-4x \\ & 8={{x}^{2}}-2x \\ & 0={{x}^{2}}-2x-8 \\ \end{align} It can be written as ${{x}^{2}}-2x-8=0$ Apply quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Here a is 1, b is $-2$ and c is $-8$ \begin{align} & x=\frac{2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -8 \right)}}{2\left( 1 \right)} \\ & =\frac{2\pm \sqrt{4+32}}{2} \\ & =\frac{2\pm \sqrt{36}}{2} \\ & =\frac{2\pm 6}{2} \end{align} On further simplification \begin{align} & x=\frac{2+6}{2} \\ & =\frac{8}{2} \\ & =4 \end{align} Or \begin{align} & x=\frac{2-6}{2} \\ & =\frac{-4}{2} \\ & =-2 \end{align} Check: Put $x=4$ in the equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$ $\begin{matrix} \frac{2\left( 4 \right)}{{{\left( 4 \right)}^{2}}+6\left( 4 \right)+8}+\frac{2}{4+2}\overset{?}{\mathop{=}}\,\frac{4}{4+4} \\ \frac{8}{16+24+8}+\frac{2}{6}\overset{?}{\mathop{=}}\,\frac{4}{8} \\ \frac{8}{48}+\frac{2}{6}\overset{?}{\mathop{=}}\,\frac{1}{2} \\ \frac{1}{6}+\frac{2}{6}\overset{?}{\mathop{=}}\,\frac{1}{2} \\ \end{matrix}$ On further simplification \begin{align} & \frac{3}{6}\overset{?}{\mathop{=}}\,\frac{1}{2} \\ & \frac{1}{2}\overset{?}{\mathop{=}}\,\frac{1}{2} \\ \end{align} Which is true. For $x=-2$ Put $x=-2$ in the equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$ \begin{align} & \frac{2\left( -2 \right)}{{{\left( -2 \right)}^{2}}+6\left( -2 \right)+8}+\frac{2}{-2+2}\overset{?}{\mathop{=}}\,\frac{-2}{-2+4} \\ & \frac{-4}{4-12+8}+\frac{2}{0}\overset{?}{\mathop{=}}\,\frac{4}{8} \end{align} Since the value of $\frac{2}{0}$ is undefined. The solution does not check for $x=-2$ Therefore, the solution set of the equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$ is $\left\{ \left. 4 \right\} \right.$.