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

Given:

Given quadratic equation is $\frac{1}{x\ -\ 3}\ +\ \frac{2}{x\ -\ 2}\ =\ \frac{8}{x};\ x\ ≠\ 0,\ 2,\ 3$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

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

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

$x(x-2+2x-6)=8(x-3)(x-2)$    (On cross multiplication)

$x(3x-8)=8(x^2-2x-3x+6)$

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

$8x^2-3x^2-40x+8x+48=0$

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

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

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

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

$x-4=0$ or $5x-12=0$

$x=4$ or $5x=12$

$x=4$ or $x=\frac{12}{5}$

The roots of the given quadratic equation are $4$ and $\frac{12}{5}$. 

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Updated on: 10-Oct-2022

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