#### Question

Solve for *x*

`1/x - 1/(x-2)=3`, x ≠ 0, 2

#### Solution

We have been given,

`1/x - 1/(x-2)=3`, x ≠ 0, 2

Now we solve the above equation as follows,

`((x-2)-x)/((x-2)(x))=3`

`(-2)/(x^2-2x)=3`

-2 = 3x^{2} - 6x

3x^{2} - 6x + 2 = 0

Now we also know that for an equation ax^{2} + bx + c = 0, the discriminant is given by the following equation:

D = b^{2} - 4ac

Now, according to the equation given to us, we have,a = 3, b = -6 and c = 2.

Therefore, the discriminant is given as,

D = (-6)^{2} - 4(3)(2)

= 36 - 24

= 12

Now, the roots of an equation is given by the following equation,

`x=(-b+-sqrtD)/(2a)`

Therefore, the roots of the equation are given as follows,

`x=(-(-6)+-sqrt12)/(2(3))`

`=(6+-2sqrt3)/6`

`=(3+-sqrt3)/3`

Now we solve both cases for the two values of *x*. So, we have,

`x=(3+sqrt3)/3`

Also,

`x=(3-sqrt3)/3`

Therefore, the value of `x=(3+sqrt3)/3`, `(3-sqrt3)/3`