Question

Solve the following quadratic equation by factorization:
x-3/x+3-x+3/x-3=48/7

Answer 1

$$\begin{gathered} \frac{\mathrm{x}-3}{\mathrm{x}+3}-\frac{\mathrm{x}+3}{\mathrm{x}-3}=\frac{48}{7} \\ \Rightarrow \frac{{\left(\mathrm{x}-3\right)}^2-{\left(\mathrm{x}+3\right)}^2}{{\mathrm{x}}^2-9}=\frac{48}{7} \\ \Rightarrow \frac{{\mathrm{x}}^2+9-6\mathrm{x}-{\mathrm{x}}^2-9-6\mathrm{x}}{{\mathrm{x}}^2-9}=\frac{48}{7} \\ \Rightarrow \frac{-12\mathrm{x}}{{\mathrm{x}}^2-9}=\frac{48}{7} \\ \Rightarrow \frac{-\mathrm{x}}{{\mathrm{x}}^2-9}=\frac{4}{7} \\ \Rightarrow 4{\mathrm{x}}^2+7\mathrm{x}-36=0 \\ \Rightarrow 4{\mathrm{x}}^2+16\mathrm{x}-9\mathrm{x}-36=0 \\ \Rightarrow 4\mathrm{x}\left(\mathrm{x}+4\right)-9\left(\mathrm{x}+4\right)=0 \\ \Rightarrow \left(\mathrm{x}+4\right)\left(4\mathrm{x}-9\right)=0 \\ \Rightarrow \mathbf{x}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{=}\frac{\mathbf{9}}{\mathbf{4}} \end{gathered}$$