Question:
Solve the following quadratic equations by factorization:
Solution:
$\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x}$
$\Rightarrow \frac{(x-2)+2(x-3)}{(x-3)(x-2)}=\frac{8}{x}$
$\Rightarrow \frac{x-2+2 x-6}{x^{2}-2 x-3 x+6}=\frac{8}{x}$
$\Rightarrow \frac{3 x-8}{x^{2}-5 x+6}=\frac{8}{x}$
$\Rightarrow x(3 x-8)=8\left(x^{2}-5 x+6\right)$
$\Rightarrow 3 x^{2}-8 x=8 x^{2}-40 x+48$
$\Rightarrow 5 x^{2}-32 x+48=0$
$\Rightarrow 5 x^{2}-20 x-12 x+48=0$
$\Rightarrow 5 x(x-4)-12(x-4)=0$
$\Rightarrow(5 x-12)(x-4)=0$
$\Rightarrow 5 x-12=0$ or $x-4=0$
$\Rightarrow x=\frac{12}{5}$ or $x=4$
Hence, the factors are 4 and $\frac{12}{5}$.
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