Question

Solve the following quadratic equation by factorization :
$\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x};x\ne 0,2,3$

• A. $\frac{1}{4}$, $\frac{12}{5}$
• B. $4$, $\frac{12}{5}$
  
• C. $\frac{-1}{4}$, $\frac{12}{5}$
• D. $-4$, $\frac{12}{5}$

Answer 1

Answer: (B)

$4$, $\frac{12}{5}$

$\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x}$

$\Rightarrow$ $\frac{x-2+2(x-3)}{(x-3)(x-2)}=\frac{8}{x}$

$\Rightarrow$ $\frac{x-2+2x-6}{x^2-2x-3x+6}=\frac{8}{x}$

$\Rightarrow$ $\frac{x+2x-8}{x^2-5x+6}=\frac{8}{x}$

$\Rightarrow$ $x^2+2x^2-8x=8x^2-40x+48$

$\Rightarrow$ $3x^2-8x=8x^2-40x+48$

$\Rightarrow$ $5x^2-32x+48=0$

$\Rightarrow$ $5x^2-20x-12x+48=0$

$\Rightarrow$ $5x(x-4)-12(x-4)=0$

$\Rightarrow$ $(x-4)(5x-12)=0$

$\Rightarrow$ $x-4=0$ and $5x-12=0$

$\therefore$ $x=4$ and $x=\frac{12}{5}$