Solve for `x` : `1/x-1/(x-2)=3,\ \ x!=0,\ 2`

Explanation

The given equation may be written as
`((x-2)-x)/(x(x-2))=3implies3x(x-2)=-2`
`implies" "3x^(2)-6x+2=0." "…(i)`
This equation is of the form `ax^(2)+bx+c=0,` where a = 3, b = -6 and c = 2.
`:." "D=(b^(2)-4ac)=(-6)^(2)-4xx3xx2=36-24=12gt0.`
So, the given equation has real roots.
Now, `sqrt(D)=sqrt(12)=2sqrt(3).`
`:." "alpha=(-b+sqrt(D))/(2a)=(6+2sqrt(3))/(2xx3)=(6+2sqrt(3))/(6)=(3+sqrt(3))/(3),`
`" "beta=(-b-sqrt(D))/(2a)=(6-2sqrt(3))/(2xx3)=(6-2sqrt(3))/(6)=(3-sqrt(3))/(3).`
Hence, the required values of x are `((3+sqrt(3)))/(3)" and "((3-sqrt(3)))/(3).`