Solve the given quadratic equation by factorization method
$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$

Open in App

Solution

We have
$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$

$\Rightarrow \frac{(x - 1) + 2 (x - 2)}{(x - 2) (x - 1)} = \frac{6}{x}$

$\Rightarrow \frac{3 x - 5}{x^{2} - 3 x + 2} = \frac{6}{x}$

$\Rightarrow 3 x^{2} - 5 x = 6 x^{2} - 18 x + 12$

$\Rightarrow 3 x^{2} - 13 x + 12 = 0 \Rightarrow 3 x^{2} - 9 x - 4 x + 12 = 0$

$\Rightarrow 3 x (x - 3) - 4 (x - 3) = 0 \Rightarrow (x - 3) (3 x - 4) = 0$

$\Rightarrow x - 3 = 0 o r 3 x - 4 = 0 \Rightarrow x = 3 o r x = \frac{4}{3}$

Suggest Corrections

Similar questions

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

\frac{x}{x + 1} + \frac{x + 1}{x} = \frac{34}{15}, x \neq 0, x \neq - 1

Solve the following quadratic equation by factorization method: [2 MARKS]

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$

x^{2} + x - (a + 1) (a + 2) = 0

x^{2} + 2 \sqrt{2} x - 6 = 0

Join BYJU'S Learning Program