How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}$$
What should I do now to get x? Can anyone show me a hint please?
 A: HINT
Divide by $\sqrt{x}$ if $x \ne 0$ to get
$$
\sqrt{x} + 1 - \sqrt{x-1} = m
$$
move $1$ to the other side and square.
A: Squaring both sides we get $$x+\sqrt{x}+x-\sqrt{x}-2\sqrt{x^2-x}=\frac{m^2x}{x+\sqrt{x}}$$
Simplifying we get
$$2x-\frac{m^2x}{x+\sqrt{x}}=2\sqrt{x^2-x}$$
simplifying we get
$$m^4x^2=-4x(x^2+x+2x\sqrt{x})$$ diviniding by $$x\neq 0$$ $$4x+4+m^4=-8\sqrt{x}$$ squaring again
$$16x^2-32x+8xm^4+8m^4+m^8+16=0$$
Can you solve this equation?
A: As you got (for $x\neq0$):
$$ x+\sqrt{x} + \sqrt{x^2-x} = m\sqrt{x} $$
$$ x+\sqrt{x} + \sqrt{x}\sqrt{x-1} = m\sqrt{x} $$
Setting $ u = \sqrt{x} $ ($ u\neq0 $) we get:
$$ u^2+u+u\sqrt{u^2-1} = mu $$
$$ u+1+\sqrt{u^2-1} = m $$
$$ \sqrt{u^2-1} = (m-1) - u$$
Squaring both sides:
$$ u^2-1 = (m-1)^2-2u(m-1)+u^2 $$
We get:
$$ u = \frac{(m-1)^2+1}{2(m-1)} $$
where $m\neq1$ and $x$ becomes:
$$ x = \Bigg(\frac{(m-1)^2+1}{2(m-1)}\Bigg)^2 $$
A: After dividing by $\sqrt{x}$ you may use the substitution $\boxed{y = \sqrt{x}}$. Then, after a bit rearranging you get
$$y+1 - m = \sqrt{y^2-1}$$
Squaring removes the $y^2$ and you can solve directly for $y$:
$$y = \frac{2-2m+m^2}{2(m-1)} \Rightarrow \boxed{x = \frac{(2-2m+m^2)^2}{4(m-1)^2}}$$
Note, that there are solutions only for $m >1$, since $y\geq 0$.
A: From the simplified form of @gt6989b, we get $$\sqrt{x}- \sqrt{x-1}=m-1.$$ 
Squaring for rationalization one must demand that $$m \ge 1...(1),~~ x \ge 1....(2),~~~\sqrt{x} \ge m-1....(3)$$.The Eq. (3) gives a very interestiong condition which has been missed out in the answers by @Dinno Koluh  and @trancelocation. When we demand (1) and (3), we get
$$\sqrt{x} \ge \frac{m^2-2m+2}{2(m-1)} \ge (m-1) \implies m^2-2m \le 0 \implies m(m-2) \le 0 \implies 1<m \le 2.$$ Finally, we clam that the solution of the priposed equation is given as
$$x=\left(\frac{m^2-2m+2}{2(m-1)}\right)^2  ~~if~~ 1<m \le 2.$$
