Question

If \(\sqrt{\dfrac{x}{1-x}}+\sqrt{\dfrac{1-x}{x}}=2\dfrac{1}{6}\) the values of \(x\) are :-

Answer 1

let one be 'a' the other one is '1/a' ... then solve for 'a'

Answer 2

I did this as follows :-

\[(\sqrt{x/1-x})^2+(\sqrt{1-x/x})^2+2(\sqrt{x/-x})(\sqrt{1-x/x})=(13/6)^2\]

Answer 3

Substitution is the key my friend.

Answer 4

not as bad as it looks i think
i added on the left

Answer 5

@experimentX will u give me the equation & @satellite73 how did u add on left ?

Answer 6

\[\frac{1}{\sqrt{x(1-x)}}=\frac{13}{6}\]

Answer 7

@hba how can we substitute it ?

Answer 8

\[ a + \frac 1 a = \frac{13} 6\]

Answer 9

LEMME TRY

Answer 10

then
\[\sqrt{x(1-x)}=\frac{6}{13}\] etc

Answer 11

X-X^2=6/13
LEFT IS COMPLETING SQUARE, RIGHT ?

Answer 12

36/169*

Answer 13

also you can take LCM, i guess you 'll get the result same as satellite did.

Answer 14

ohh ! i gt it now thanx :)

Answer 15

i bet there is a snazzier way, i just grind it till i find it, but i would like to see snapper solution