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Solve the following quadratic equation by factorization:
$\frac{1}{x\ -\ 2}\ +\ \frac{2}{x\ -\ 1}\ =\ \frac{6}{x},\ x\ ≠\ 0$
Given:
Given quadratic equation is $\frac{1}{x\ -\ 2}\ +\ \frac{2}{x\ -\ 1}\ =\ \frac{6}{x},\ x\ ≠\ 0$.
To do:
We have to solve the given quadratic equation by factorization.
Solution:
$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$
$\frac{1(x-1)+(2)(x-2)}{(x-2)(x-1)}=\frac{6}{x}$
$\frac{x-1+2x-4}{x^2-x-2x+2}=\frac{6}{x}$
$\frac{3x-5}{x^2-3x+2}=\frac{6}{x}$
$x(3x-5)=6(x^2-3x+2)$ (on cross multiplication)
$3x^2-5x=6x^2-18x+12$
$(6-3)x^2+(-18+5)x+12=0$
$3x^2-13x+12=0$
$3x^2-9x-4x+12=0$ ($9+4=13$ and $9\times4=3\times12$)
$3x(x-3)-4(x-3)=0$
$(3x-4)(x-3)=0$
$3x-4=0$ or $x-3=0$
$3x=4$ or $x=3$
$x=\frac{4}{3}$ or $x=3$
The roots of the given quadratic equation are $\frac{4}{3}$ and $3$. 
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