Solve the following quadratic equations by factorization:

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}, x \neq 0$

Open in App

Solution

We have been given

Therefore,

Therefore,

or,

Hence, or .

Suggest Corrections

Similar questions

Q. Solve the following quadratic equations by factorization:

\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}

Solve the following quadratic equation by factorization.
$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$

Q. Solve the following quadratic equations by factorization:

\frac{2}{x + 1} + \frac{3}{2 (x - 2)} = \frac{23}{5 x}; x \neq 0, - 1, 2

Q. Solve the following quadratic equation by factorization and find its root :

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

Solve the following quadratic equation by factorization.
$3 x^{2} - 2 \sqrt{6 x} + 2 = 0$

Join BYJU'S Learning Program

Solve the following quadratic equations by factorization: 1x-2+2x-1=6x, x≠0

We have been given Therefore, Therefore, or, Hence, or .

Solve the following quadratic equations by factorization:

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}, x \neq 0$

We have been given

Therefore,

Therefore,

or,

Hence, or .