Question

Solve the following quadratic equations by factorization:

$\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x};x\ne 0,2,3$

Answer 1

$$\begin{gathered} \frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x} \\ \Rightarrow \frac{\left(x-2\right)+2\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=\frac{8}{x} \\ \Rightarrow \frac{x-2+2x-6}{x^2-2x-3x+6}=\frac{8}{x} \\ \Rightarrow \frac{3x-8}{x^2-5x+6}=\frac{8}{x} \\ \Rightarrow x\left(3x-8\right)=8\left(x^2-5x+6\right) \\ \Rightarrow 3x^2-8x=8x^2-40x+48 \\ \Rightarrow 5x^2-32x+48=0 \\ \Rightarrow 5x^2-20x-12x+48=0 \\ \Rightarrow 5x\left(x-4\right)-12\left(x-4\right)=0 \\ \Rightarrow \left(5x-12\right)\left(x-4\right)=0 \\ \Rightarrow 5x-12=0\mathrm{or}x-4=0 \\ \Rightarrow x=\frac{12}{5}\mathrm{or}x=4 \end{gathered}$$

Hence, the factors are 4 and $\frac{12}{5}$.