Question

Solve $\frac{2}{x^2}-\frac{5}{x}+2=0$ by factorisation method. Find the roots of (PS)
$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}x\ne -4,7$

Answer 1

Given,

$\frac{2}{x^2}-\frac{5}{x}+2=0$

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

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

$(x-2)(2x-1)=0$

$x=2,\frac{1}{2}$

Given,

$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$

$30(x-7)-30(x+4)=11(x+4)(x-7)$

$30x-210-30x-120=11x^2-33x-308$

$11x^2-33x+22=0$

$x=\frac{-(-33)\pm \sqrt{{(-33)}^2-4\cdot 11\cdot 22}}{2\cdot 11}$

$=\frac{33\pm \sqrt{121}}{2\cdot 11}$

$=\frac{33\pm 11}{22}$

$\therefore x=2,1$