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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Updated on: 10-Oct-2022

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