Question

`x-1/x=3,x≠0` 

Answer 1

The given equation is 

`x-1/x=3,x≠0` 

⇒`(x^2-1)/x=3` 

⇒`x^2-1=3x` 

⇒`x^2-3x-1=0` 

This equation is of the form `ax^2+bx+c=0` where` a=1,b=-3 and c=-1` 

∴ Discriminant, `D=b^2-4ac=(-3)^2-4xx1xx(-1)=9+4=13>0` 

So, the given equation has real roots.
Now, `sqrtD=sqrt13` 

∴ `α=(-b+sqrt(D))/(2a)=(-(-3)+sqrt(13))/(2xx1)=(3+sqrt(13))/2`

β=`(-b+sqrt(D))/(2a)=(-(-3)+sqrt(13))/(2xx1)=(3+sqrt(13))/2=(3-sqrt(13))/2` 

Hence, `(3+sqrt(13))/3` and `(3-sqrt(13))/3`are the roots of the given equation.