# Find the roots of quadratic equation 1/x - 1/x-2 = 3.

Answer: $x = \frac{3 + \sqrt{3}}{3}$ and $x = \frac{3 – \sqrt{3}}{3}$.

$\frac{1}{x}-\frac{1}{x-2}=3$

Simplify the above equation we get

$\frac{()x – 2 – x)}{x(x-2)}=3$

3x(x – 2) = x – 2 – x

3x2 – 6x = -2

3x2 – 6x + 2 = 0

The above equation is in the form of ax2 + bx + c = 0

Where

a = 3

b = -6

c = 2

So value of x can be written as

$x = \frac{-b\pm \sqrt{b^{2}}-4ac}{2a}$

Put the values of a, b and c in the above equation we get

$x = \frac{-(-6)\pm \sqrt{(-6^{2}})-4\times 3\times 2}{2\times 3}$ $x = \frac{6\pm \sqrt{36 -24}}{6}$ $x = \frac{6\pm \sqrt{12}}{6}$ $x = \frac{6\pm 2\sqrt{3}}{6}$ $x = \frac{3\pm \sqrt{3}}{3}$

Hence value of x is either

$x = \frac{3 + \sqrt{3}}{3}$ or

$x = \frac{3 – \sqrt{3}}{3}$

∴ the roots of the quadratic equation $\frac{1}{x}-\frac{1}{x-2}=3$ is

$x = \frac{3 + \sqrt{3}}{3}$ and $x = \frac{3 – \sqrt{3}}{3}$.