Question

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

 

Answer 1

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

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

${\displaystyle 2x^2-5x+2=0}$ 

${\displaystyle 4x^2-10x+4=0}$                      (Multiplying both sides by 2) 

${\displaystyle 4x^2-10x=-4}$ 

${\displaystyle {\left(2x\right)}^2-2\times 2x\times \frac{5}{2}+{\left(\frac{5}{2}\right)}^2=-4+{\left(\frac{5}{2}\right)}^2}$       [Adding ${\displaystyle {\left(\frac{5}{2}\right)}^2}$ on both sides ] 

${\displaystyle {(2x-\frac{5}{2})}^2=-4+\frac{25}{4}=\frac{9}{4}={\left(\frac{3}{2}\right)}^2}$ 

${\displaystyle 2x-\frac{5}{2}=\pm \frac{3}{2}}$                        (Taking square root on both sides) 

${\displaystyle 2x-\frac{5}{2}=\frac{3}{2} \text{or} 2x-\frac{5}{2}=-\frac{3}{2}}$ 

${\displaystyle 2x=\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4}$  or  ${\displaystyle 2x=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1}$ 

${\displaystyle x=2 \text{or} x=\frac{1}{2}}$ 

Hence, 2 and ${\displaystyle \frac{1}{2}}$ are the roots of the given equation.