# How do you solve (4x-2)/(x-6) = -x/(x+5) and find any extraneous solutions?

Jul 25, 2016

The solns. are ${x}_{1} \cong 0.655 , \mathmr{and} , {x}_{2} \cong - 3.055$.

#### Explanation:

$\frac{4 x - 2}{x - 6} = - \frac{x}{x + 5}$

By Cross-Multiplication, $\left(4 x - 2\right) \left(x + 5\right) = - x \left(x - 6\right)$

By Expansion, $4 {x}^{2} + 20 x - 2 x - 10 = - {x}^{2} + 6 x$

By Transposition, $4 {x}^{2} + 18 x - 10 + {x}^{2} - 6 x = 0$

$\therefore 5 {x}^{2} + 12 x - 10 = 0. \ldots \ldots \ldots \ldots \left(1\right)$

We use the Formula to find the roots of this Qudr. Eqn. :-

The roots $\alpha , \beta$ of the Gen. Qudr. Eqn. $a {x}^{2} + b x + c = 0$ are

$\alpha = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} , \beta = \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}$

In our case, since, $a = 5 , b = 12 , c = - 10$, we get,

$\alpha = \frac{- 12 + \sqrt{{12}^{2} - 4 \cdot 5 \cdot \left(- 10\right)}}{2 \cdot 5} = \frac{- 12 + \sqrt{144 + 200}}{10}$

$\therefore \alpha = \frac{- 12 + \sqrt{344}}{10} , \mathmr{and} , \beta = \frac{- 12 - \sqrt{344}}{10}$

Taking, $\sqrt{344} \cong 18.55 , \alpha \cong 0.655 , \mathmr{and} , \beta \cong - 3.055$

Thus, the solns. are ${x}_{1} = \alpha \cong 0.655 , \mathmr{and} , {x}_{2} = \beta \cong - 3.055$.