Solve the following quadratic equation using quadratic formula: $\frac{x - 1}{x + 1} + \frac{x - 3}{x - 4} = \frac{10}{3}$

{\frac{(9 + \sqrt{769})}{8}, \frac{(9 - \sqrt{769})}{8}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{\frac{(9 - \sqrt{769})}{8}, \frac{(9 - \sqrt{769})}{8}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{\frac{(9 + \sqrt{769})}{8}, \frac{(9 + \sqrt{769})}{8}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A ${\frac{(9 + \sqrt{769})}{8}, \frac{(9 - \sqrt{769})}{8}}$
Given equation is $\frac{x - 1}{x + 1} + \frac{x - 3}{x - 4} = \frac{10}{3}$

$\Rightarrow$ $\frac{(x - 1) (x - 4) + (x - 3) (x + 1)}{(x + 1) (x - 4)} = \frac{10}{3}$

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

$\Rightarrow$ $\frac{2 x^{2} - 7 x + 1}{x^{2} - 3 x - 4} = \frac{10}{3}$

$\Rightarrow$ $6 x^{2} - 21 x + 3 = 10 x^{2} - 30 x - 40$

$\Rightarrow$ $4 x^{2} - 9 x - 43 = 0$

$\Rightarrow$ $a = 4, b = - 9, c = - 43$

$\Rightarrow$ $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{- (- 9) \pm \sqrt{(- 9)^{2} - 4 (4) (- 43)}}{2 (4)}$

$= \frac{9 \pm \sqrt{769}}{8}$

$∴$ $x = \frac{9 + \sqrt{769}}{8}$ and $x = \frac{9 - \sqrt{769}}{8}$

Suggest Corrections