Question

12.) x-1/x-2+x-3/x-4=10/3
pls answer this question if u know the solution <3​

Answer 1

Given:

      $x - \frac{1}{x} - 2 + x - \frac{3}{x} - 4 = \frac{10}{3}$

To Find:

     x

Step-by-step explanation:

•   $x - \frac{1}{x} - 2 + x - \frac{3}{x} - 4 = \frac{10}{3}$

• $x - \frac{1}{x} + x - \frac{3}{x} - 6 = \frac{10}{3}$

• $2x - \frac{1}{x} - \frac{3}{x} - 6 = \frac{10}{3}$

• $2x - \frac{1-3}{x} = \frac{10}{3} + 6$

• $2x - \frac{-2}{x} = \frac{10}{3} + \frac{18}{3}$

• $\frac{2x^{2} }{x} + \frac{2}{x} = \frac{28}{3}$

• $\frac{2x^{2} + 2}{x} = \frac{28}{3}$

• $2x^{2} + 2 = \frac{28 \times x}{3}$

• $2(x^{2} + 1) = \frac{28 \times x}{3}$

• $3(x^{2} + 1) = \frac{28 \times x}{2}$

• $3x^{2} + 3 = 14 \times x$

• $3x^{2} + 3 = 14x$

• $3x^{2} - 14x + 3 = 0$

• This equation can be represented as a$x^{2}$ + bx +c = 0

• a = 3, b = -14 and c =3

• Now, using Shree Dharacharya Formula for x

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

• putting the value in it

• $x = \frac { - (-14) \pm \sqrt{ (-14)^2 - 4 \times 3 \times 3} } { 2\times 3}$

• $x = \frac { 14 \pm \sqrt{ (196 - 36} } { 6}$

$x = \frac { 14 \pm \sqrt{ (160} } { 6}$

Solving this we get two values of x.