Question

`2/x^2-5/x+2=0` 

 

Answer 1

`2/x^2-5/x+2=0`  

`(2-5x+2x^2)/x^2=0` 

`2x^2-5x+2=0` 

`4x^2-10x+4=0`                      (Multiplying both sides by 2) 

`4x^2-10x=-4` 

`(2x)^2-2xx2x xx5/2+(5/2)^2=-4+(5/2)^2`       [Adding `(5/2)^2` on both sides ] 

`(2x-5/2)^2=-4+25/4=9/4=(3/2)^2` 

`2x-5/2=+-3/2`                        (Taking square root on both sides) 

`2x-5/2=3/2  or  2x-5/2=-3/2` 

`2x=3/2+5/2=8/2=4`  or  `2x=-3/2+5/2=2/2=1` 

`x=2  or  x=1/2` 

Hence, 2 and `1/2` are the roots of the given equation.