Question

Solve the following quadratic equation by factorization :
$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3};x\ne 5,7$

The roots are $8,\frac{11}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}$

$\Rightarrow$ $\frac{(x-4)(x-7)+(x-6)(x-5)}{(x-5)(x-7)}=\frac{10}{3}$

$\Rightarrow$ $\frac{x^2-7x-4x+28+x^2-5x-6x+30}{x^2-7x-5x+35}=\frac{10}{3}$

$\Rightarrow$ $\frac{2x^2-22x+58}{x^2-12x+35}=\frac{10}{3}$

$\Rightarrow$ $\frac{2(x^2-11x+29)}{x^2-12x+35}=\frac{10}{3}$

$\Rightarrow$ $\frac{x^2-11x+29}{x^2-12x+35}=\frac{5}{3}$

$\Rightarrow$ $3x^2-33x+87=5x^2-60x+175$

$\Rightarrow$ $2x^2-27x+88=0$

$\Rightarrow$ $2x^2-16x-11x+88=0$

$\Rightarrow$ $2x(x-8)-11(x-8)=0$

$\Rightarrow$ $(x-8)(2x-11)=0$

$\Rightarrow$ $x=8=0$ and $2x-11=0$

$\therefore$ $x=8$ and $x=\frac{11}{2}$

$\therefore$ Roots given in question are correct.