Question

Solve the following quadratic equation by factorization, the roots are :$6$, $\frac{40}{13}$
$\frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

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

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

$\Rightarrow$ $13x^2-78x-40x+240=0$

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

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

$\Rightarrow$ $x-6=0$ and $13x-40=0$

$\therefore$ $x=6$ and $x=\frac{40}{13}$

$\therefore$ We can see roots given in question are correct.