# Question 51c56

##### 2 Answers
Aug 17, 2017

$16 {x}^{4} + 232 {x}^{3} + 673 {x}^{2} - 818 x + 41 = 0$

#### Explanation:

sqrtx div sqrt(1-x)+sqrt(1-x)=5÷2#

$\Rightarrow \frac{\sqrt{x}}{\sqrt{1 - x}} + \sqrt{1 - x} = \frac{5}{2}$

Taking L.C.M

$\Rightarrow \frac{\sqrt{x} + 1 - x}{\sqrt{1 - x}} = \frac{5}{2}$

Cross Multiply

$\Rightarrow 2 \left(\sqrt{x} + 1 - x\right) = 5 \sqrt{1 - x}$

$\Rightarrow 2 \sqrt{x} + 2 - 2 x = 5 \sqrt{1 - x}$

Collect Like Terms

$\Rightarrow 2 \sqrt{x} - 5 \sqrt{1 - x} = 2 x + 2$

Square both sides

$\Rightarrow {\left(2 \sqrt{x} - 5 \sqrt{1 - x}\right)}^{2} = {\left(2 x + 2\right)}^{2}$

$\Rightarrow \left(2 \sqrt{x} - 5 \sqrt{1 - x}\right) \left(2 \sqrt{x} - 5 \sqrt{1 - x}\right) = \left(2 x + 2\right) \left(2 x + 2\right)$

$\Rightarrow 4 \left(x\right) - 20 \sqrt{1 - x} + 25 \left(1 - x\right) = 4 {x}^{2} + 8 x + 4$

$\Rightarrow 4 \left(x\right) - 20 \sqrt{1 - x} + 25 - 25 x = 4 {x}^{2} + 8 x + 4$

$\Rightarrow - 20 \sqrt{1 - x} + 25 - 21 x = 4 {x}^{2} + 8 x + 4$

$\Rightarrow - 20 \sqrt{1 - x} = 4 {x}^{2} + 8 x + 4 + 21 x - 25$

$\Rightarrow - 20 \sqrt{1 - x} = 4 {x}^{2} + 29 x - 21$

Square both sides

$\Rightarrow {\left(- 20 \sqrt{1 - x}\right)}^{2} = {\left(4 {x}^{2} + 29 x - 21\right)}^{2}$

$\Rightarrow 400 \left(1 - x\right) = {\left(4 {x}^{2} + 29 x - 21\right)}^{2}$

$\Rightarrow 400 - 400 x = \left(4 {x}^{2} + 29 x - 21\right) \left(4 {x}^{2} + 29 x - 21\right)$

$\Rightarrow 400 - 400 x = 16 {x}^{4} + 232 {x}^{3} + 673 {x}^{2} - 1218 x + 441$

Collect like terms

$\Rightarrow 16 {x}^{4} + 232 {x}^{3} + 673 {x}^{2} - 1218 x + 400 x + 441 - 400 = 0$

$\Rightarrow 16 {x}^{4} + 232 {x}^{3} + 673 {x}^{2} - 818 x + 41 = 0$

Solve the polynomial above..

That's how far i could get, But in my own point of view, i strongly doubt the Authenticity of the question..

Aug 17, 2017

See below.

#### Explanation:

$\frac{\sqrt{x}}{\sqrt{1 - x}} + \sqrt{1 - x} = 2.5$ Calling $y = 1 - x$ we have

$\sqrt{1 - y} + y = 2.5 \sqrt{y}$ or

$1 - y = {\left(2.5 \sqrt{y} - y\right)}^{2}$ or

$1 - y = {2.5}^{2} y - 5 y \sqrt{y} + {y}^{2}$ and now making $\xi = \sqrt{y}$ we have

${\xi}^{4} - 5 {\xi}^{3} + \left(1 + {2.5}^{2}\right) {\xi}^{2} - 1 = 0$ This polynomial has two real roots

$\xi = \left\{- 0.332869 , 0.43602\right\}$ giving

$y = \left\{0.110802 , 0.190114\right\}$ and consequently

$x = \left\{0.889198 , 0.809886\right\}$

but substituting into the original equation only $x = 0.809886$ is feasible, so the solution is

$x = 0.809886$