Question

Solve the following quadratic equation by factorization :
$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3};x\ne 3,5$,

The roots are 6, $\frac{7}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}$

$\Rightarrow$ $\frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}$

$\Rightarrow$ $\frac{x^2-5x-2x+10+x^2-3x-4x+12}{x^2-5x-3x+15}=\frac{10}{3}$

$\Rightarrow$ $\frac{2x^2-14x+22}{x^2-8x+15}=\frac{10}{3}$

$\Rightarrow$ $\frac{2(x^2-7x+11)}{x^2-8x+15}=\frac{10}{3}$

$\Rightarrow$ $\frac{x^2-7x+11}{x^2-8x+15}=\frac{5}{3}$

$\Rightarrow$ $3x^2-21x+33=5x^2-40x+75$

$\Rightarrow$ $2x^2-19x+42=0$

$\Rightarrow$ $2x^2-12x-7x+42=0$

$\Rightarrow$ $2x(x-6)-7(x-6)=0$

$\Rightarrow$ $(x-6)(2x-7)=0$

$\Rightarrow$ $x-6=0$ and $2x-7=0$

$\Rightarrow$ $x=6$ and $x=\frac{7}{2}$

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