Question

Solve: $\frac{x}{1-x}+\sqrt{\frac{1-x}{x} } = 2\frac{1}{6} ,x\neq 0\:\:and\:\:x\neq 1.$

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Answer 1

Correct Question :

Solve :

$\bf{\sqrt{\dfrac{x}{1 - x}} + \sqrt{\dfrac{1 - x}{x}} = 2\dfrac{1}{6}}$

Solution :

Let $\bf{\sqrt{\dfrac{a}{1 - a}}}$ be z.

So, $\bf{\sqrt{\dfrac{1 - a}{a}}}$ will be 1/z.

So by replacing the values (in terms of z) of $\bf{\sqrt{\dfrac{a}{1 - a}}}$ and $\bf{\sqrt{\dfrac{1 - a}{a}}}$ , we get the new Equation as :

$:\implies \bf{z + \dfrac{1}{z} = 2\dfrac{1}{6}}$

$:\implies \bf{z + \dfrac{1}{z} = \dfrac{(6 \times 2) + 1}{6}}$

$:\implies \bf{z + \dfrac{1}{z} = \dfrac{12 + 1}{6}}$

$:\implies \bf{z + \dfrac{1}{z} = \dfrac{13}{6}}$

$:\implies \bf{\dfrac{z^{2} + 1}{z} = \dfrac{13}{6}}$

By cross-multiplication , we get :

$:\implies \bf{6(z^{2} + 1) = 13 \times z}$

$:\implies \bf{6z^{2} + 6 = 13z}$

$:\implies \bf{6z^{2} - 13z + 6 = 0}$

$\boxed{\therefore \bf{6z^{2} - 13z + 6 = 0}}$

Hence, here the equation is formed i.e,

$\bf{6z^{2} - 13z + 6 = 0}$

Now by solving it by middle-splitting factor theorem, we get :

$:\implies \bf{6z^{2} - 13z + 6 = 0}$

$:\implies \bf{6z^{2} - (9 + 4)z + 6 = 0}$

$:\implies \bf{6z^{2} - 9 - 4z + 6 = 0}$

$:\implies \bf{3z(2z - 3) - 2(2z -3) = 0}$

$:\implies \bf{(2z - 3)(3z - 2) = 0}$

$:\implies \bf{(2z - 3 = 0)\:;\:(3z - 2 = 0)}$

$:\implies \bf{(2z = 3)\:;\:(3z = 2)}$

$:\implies \bf{z = \dfrac{3}{2}\:;\:z = \dfrac{2}{3}}$

$\boxed{\therefore \bf{z = \dfrac{3}{2}\:;\:z = \dfrac{2}{3}}}$

Hence, the value of $\bf{\sqrt{\dfrac{a}{1 - a}}}$ is $\bf{\dfrac{2}{3}}$ or $\bf{\dfrac{3}{2}}$.

Here , we gave got two values of $\bf{\sqrt{\dfrac{a}{1 - a}}}$ , so by putting these value in the Equation $\bf{\sqrt{\dfrac{a}{1 - a}}}$, we get :

By putting the value of z as 3/2 in the Equation , we get :

$:\implies \bf{\sqrt{\dfrac{a}{1 - a}} = \dfrac{3}{2}}$

By Squaring on both the sides , we get :

$:\implies \bf{\bigg(\sqrt{\dfrac{a}{1 - a}}\bigg)^{2} = \bigg(\dfrac{3}{2}\bigg)^{2}}$

$:\implies \bf{\dfrac{a}{1 - a} = \dfrac{9}{4}}$

By Cross-multiplication , we get :

$:\implies \bf{4a = 9(1 - a)}$

$:\implies \bf{4a = 9 - 9a}$

$:\implies \bf{4a + 9a = 9}$

$:\implies \bf{13a = 9}$

$:\implies \bf{a = \dfrac{9}{13}}$

$\boxed{\therefore \bf{a = \dfrac{9}{13}}}$

Hence the value of a is 9/13.

By putting the value of z as 2/3 in the Equation , we get :

$:\implies \bf{\sqrt{\dfrac{a}{1 - a}} = \dfrac{2}{3}}$

By Squaring on both the sides , we get :

$:\implies \bf{\bigg(\sqrt{\dfrac{a}{1 - a}}\bigg)^{2} = \bigg(\dfrac{2}{3}\bigg)^{2}}$

$:\implies \bf{\dfrac{a}{1 - a} = \dfrac{4}{9}}$

By Cross-multiplication , we get :

$:\implies \bf{9a = 4(1 - a)}$

$:\implies \bf{9a = 4 - 4a}$

$:\implies \bf{9a + 4a = 4}$

$:\implies \bf{13a = 4}$

$:\implies \bf{a = \dfrac{4}{13}}$

$\boxed{\therefore \bf{a = \dfrac{4}{13}}}$

Hence the value of a is 4/13.

Thus the value of a is 9/13 and 4/13.

Answer 2

Answer:

Step-by-step explanation:

$\Huge\boxed{Hey\:friend\:here\:your\:answer}$:

Correct question:

$\large\boxed{\sqrt{\frac{x}{1-x}}+ \sqrt{\frac{1-x}{x} } = 2\frac{1}{6} ,x\neq 0\:and\:x\neq 1.}$

∴Given equation reduces to:

   $\large\boxed{y+\frac{1}{y}=\frac{13}{6}}$  

$\small\boxed{6y^2+6=13y},\\\\\small\boxed{6y^2=13y+6=0},\\\\\small\boxed{(2y-3)(3y-2)=0}\\\\\small\boxed{y=\frac{3}{2} , or\:y=\frac{2}{3}}$

When:$\large\boxed{y=\frac{3}{2} = \sqrt{\frac{x}{1-x} } =\frac{3}{2} =\frac{x}{1-x}=\frac{9}{4}}$,

→$\small\boxed{4x=9-9x = x= \frac{9}{13}}$,

And:$\large\boxed{y=\frac{2}{3}=\sqrt{\frac{x}{1-x} } =\frac{2}{3}=\frac{x}{1-x} =\frac{4}{9}}$,

→$\small\boxed{9x=4-4x = x=\frac{4}{13}},\\\\\huge\boxed{So,\:required\:solution\:is:x=\frac{9}{13} ,\:or\:\frac{4}{13}}$

$\huge\boxed{Hope\:it\:helps\:you}$