Solve the following quadratic equation by factorization:
$\frac{2x}{x\ -\ 4}\ +\ \frac{2x\ -\ 5}{x\ -\ 3}\ =\ \frac{25}{3},\ x\ ≠\ 3,\ 4$

Given:

Given quadratic equation is $\frac{2x}{x\ -\ 4}\ +\ \frac{2x\ -\ 5}{x\ -\ 3}\ =\ \frac{25}{3},\ x\ ≠\ 3,\ 4$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}$

$\frac{2x(x-3)+(2x-5)(x-4)}{(x-4)(x-3)}=\frac{25}{3}$

$\frac{2x^2-6x+2x^2-8x-5x+20}{x^2-3x-4x+12}=\frac{25}{3}$

$\frac{4x^2-19x+20}{x^2-7x+12}=\frac{25}{3}$

$3(4x^2-19x+20)=25(x^2-7x+12)$

$12x^2-57x+60=25x^2-175x+300$

$(25-12)x^2+(-175+57)x+300-60=0$

$13x^2-118x+240=0$

$13x^2-78x-40x+240=0$   ($78+40=118$ and $78\times40=13\times240$)

$13x(x-6)-40(x-6)=0$

$(13x-40)(x-6)=0$

$13x-40=0$ or $x-6=0$

$13x=40$ or $x=6$

$x=\frac{40}{13}$ or $x=6$

The roots of the given quadratic equation are $\frac{40}{13}$ and $6$.

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Updated on: 10-Oct-2022

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