Question

under root x upon 1 - X + under root 1 - X upon X equals to 13 upon 6​

Answer 1

Answer:

$\bf\huge\red{x =\frac{4 ±\sqrt{7}}{3}}$

Step-by-step explanation:

It is being given that,

$\frac{ \sqrt{x} }{1 - x} + \frac{1 - x}{ \sqrt{x} } = \frac{13}{6}$

Now, taking the LCM and solving,

we get,

$= > \frac{ { (\sqrt{x} )}^{2} + {(1 - x)}^{2} }{ \sqrt{x} (1 - x)} = \frac{13}{6}$

Further ,

$= > \frac{x + 1 + {x}^{2} - 2x}{ \sqrt{x}(1 - x) } = \frac{13}{6}$

$= > {6( {x}^{2} - x + 1)} = 13 \sqrt{x} (1 - x)$

Let, us take

$\frac{ \sqrt{x} }{1 - x} = a$

So, we get

$= > a + \frac{1}{a} = \frac{13}{6}$

$= > \frac{ {a}^{2} + 1 }{a } = \frac{13}{6}$

On cross multiplication, we get

$= > 6( {a}^{2} + 1) = 13a$

$= > 6 {a}^{2} - 13a + 6 = 0$

After solving the values of a, we get

$= > a = \frac{13 + \sqrt{169 - 144} }{12}$

$= > a = \frac{13 + \sqrt{25} }{12 } \\ \\ = > a = \frac{13 + 5}{12} \\ \\ = > a = \frac{18}{12} \\ \\ = > a = \frac{3}{2}$

Therefore, we get

$= > \frac{ \sqrt{x} }{1 - x} = \frac{3}{2}$

Now, squaring both sides and doing cross multiplication,

we get,

$= > 2x = 3( {x}^{2} - 2x + 1)$

$= > 3 {x}^{2} - 8x + 3 = 0$

Now, solving for x, we get

$= > x = \frac{8 ± \sqrt{64 - 36} }{6} \\ \\ = > x = \frac{8 ± \sqrt{28} }{6} \\ \\ = > x = \frac{8 ± 2 \sqrt{7} }{6} \\ \\ = > x = \frac{4 ± \sqrt{7} }{3}$