Question #51c56

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Question #51c56

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Answer 1 · 2017-08-17T10:05:27

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

Explanation:

$\sqrt{x} \div \sqrt{1 - x} + \sqrt{1 - x} = 5 \div 2$ $sqrtx div sqrt(1-x)+sqrt(1-x)=5÷2$

$\Rightarrow \frac{\sqrt{x}}{\sqrt{1 - x}} + \sqrt{1 - x} = \frac{5}{2}$ $rArr sqrtx/sqrt(1 - x) + sqrt (1 - x) = 5/2$

Taking L.C.M

$\Rightarrow \frac{\sqrt{x} + 1 - x}{\sqrt{1 - x}} = \frac{5}{2}$ $rArr (sqrtx + 1 - x)/sqrt(1 - x) = 5/2$

Cross Multiply

$\Rightarrow 2 (\sqrt{x} + 1 - x) = 5 \sqrt{1 - x}$ $rArr 2 (sqrtx + 1 - x) = 5 sqrt(1 - x)$

$\Rightarrow 2 \sqrt{x} + 2 - 2 x = 5 \sqrt{1 - x}$ $rArr 2sqrtx + 2 - 2x = 5 sqrt(1 - x)$

Collect Like Terms

$\Rightarrow 2 \sqrt{x} - 5 \sqrt{1 - x} = 2 x + 2$ $rArr 2sqrtx - 5 sqrt(1 - x) = 2x + 2$

Square both sides

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

$\Rightarrow (2 \sqrt{x} - 5 \sqrt{1 - x}) (2 \sqrt{x} - 5 \sqrt{1 - x}) = (2 x + 2) (2 x + 2)$ $rArr (2sqrtx - 5 sqrt(1 - x)) (2sqrtx - 5 sqrt(1 - x)) = (2x + 2) (2x + 2)$

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

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

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

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

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

Square both sides

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

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

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

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

Collect like terms

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

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

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..

Answer 2 · 2017-08-17T10:50:16

See below.

Explanation:

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

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

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

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

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

$ξ = {- 0.332869, 0.43602}$ $xi = {-0.332869,0.43602}$ giving

$y = {0.110802, 0.190114}$ $y = {0.110802,0.190114}$ and consequently

$x = {0.889198, 0.809886}$ $x = {0.889198,0.809886}$

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

$x = 0.809886$ $x = 0.809886$

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Question #51c56

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