Question

HEY FRIEND PLS... ANSWER ME. ITS URGENT.

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

Answer 1

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

= > (x - 1)(x + 4) + (x - 3)(x - 2) = (x - 2)(x + 4)

= > x^2 + 3x - 4 + x^2 - 5x + 6 = x^2 + 2x - 8

= > 2x^2 - 2x + 2 = x^2 + 2x - 8

= > 2x^2 - 4x + 10 = x^2

= > x^2 - 4x + 10 = 0

It is in the form of ax^2 + bx + c = 0, we get

a = 1 , b = -4, c = 10

(1)

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

$= \ \textgreater \ \frac{-(-4) + \sqrt{(-4)^2 - 4 * 10} }{2 * 1}$

$= \ \textgreater \ \frac{4 + \sqrt{24i} }{2}$

$= \ \textgreater \ \frac{2(2 + \sqrt{6}i) }{2}$

$= \ \textgreater \ 2 + \sqrt{6} i$



(2)

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

$= \ \textgreater \ \frac{-(-4) - \sqrt{(-4)^2 - 4 * 1 * 10} }{2}$

$= \ \textgreater \ \frac{4 - \sqrt{24} }{2}$

$= \ \textgreater \ 2 - \sqrt{6} i$




Hope this helps!

Answer 2

Hi,

Please see the attached file!


Thanks