Question

`1/x-1/(x-2)=3,x≠0,2` 

Answer 1

The given equation is 

`1/x-1/x-2=3,x≠0,2` 

⇒   `(x-2-x)/(x(x-2))=3` 

⇒`-2/(x^2-2x)=3` 

⇒`-2=3x^2-6x` 

⇒`3x^2-6x+2=0` 

This equation is of the form `ax^2+bx+c=0` Where `a=3``, b=-6` and `c=2` 

∴Discriminant, `D=b^2-4ac=(-6)^2-4xx3xx2=36-24=12>0` 

So, the given equation has real roots.
Now, `sqrtD=sqrt12=2sqrt3` 

∴ `α=(-b+sqrt(D))/(2a)=(-(-6)+2sqrt(3))/(2xx3)=(6+2sqrt(3))/6=(3+sqrt(3))/3` 

`β= (-b-sqrt(D))/(2a)=(-(-6)-2sqrt(3))/(2xx3)=(6-2sqrt(3))/6=(3-sqrt(3))/3`  

Hence, `(3+sqrt3)/3`  and`(3-sqrt3)/3` are the roots of the given equation.