Question

1/x - 1/(x-2) = 3 Solve by Factorization

Answer 1

this can be solved by discriminant method

Answer 2

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

$= > \frac{(x - 2) - x}{x(x - 2)} =3$

$= > \frac{x - 2 - x}{x(x - 2)} =3$

$= > \frac{-2}{x(x - 2)}=3$

⇒ -2 = 3x(x - 2)

⇒ -2 = 3x^2 - 6x

⇒ 3x^2 - 6x + 2 = 0.

On comparing with ax^2 + bx + c = 0, we get a = 3, b = -6, c = 2.

The solutions are:

(i)

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

$= > \frac{6 + \sqrt{(-6)^2 - 4(3)(2)}}{2 * 3}$

$= > \frac{6 + \sqrt{36 - 24}}{6}$

$= > \frac{6 + \sqrt{12}}{6}$

$= > \frac{6 + 2\sqrt{3}}{6}$

$= > \frac{2(3 + \sqrt{3})}{2 * 3}$

$= > \frac{3+\sqrt{3}}{3}$

(ii)

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

$= > \frac{6 - \sqrt{(-6)^2 - 4(3)(2)}}{2 * 3}$

$= > \frac{6 - \sqrt{12}}{6}$

$= > \frac{6 - 2\sqrt{3}}{6}$

$= > \frac{2(3 - \sqrt{3})}{6}$

$= > \frac{3 - \sqrt{3}}{3}$

Therefore, the solutions are:

$= > \boxed{x = \frac{3 + \sqrt{3}}{3}, \frac{3 - \sqrt{3}}{3}}$

Note: It cannot be solved using Factorization.

Hope this helps!