Solving a parametric equation I'm trying to solve the following equation:
$\sqrt{m+\sqrt{x^{2}+m^{2}}}=x$.
but I'm clueless. Could you help me? 
I tried the following steps: $m+\sqrt{x^{2}+m^{2}}=x^{2} $
$x^{2}+m^{2}=x^{4}-2x^{2}m+m^{2}$
$x^{2}=x^{4}-2x^{2}m$
$x^{2}*(1-x^{2})=x^{2}*(-2m)$
$1-x^{2}=-2m$
$x=\sqrt{2m+1}$
$x=\sqrt{3}$ for m=0, $x\in[\sqrt{3},\infty)$ for m>0 and $x\in[0,\sqrt{3}]$ for m<0.
Are my steps correct? Thanks
 A: Up to $x=\sqrt{2m+1}$ this is fine
You should write explicitly that $x\ge 0$ from the original equation to be thorough (else $-\sqrt{2m+1}$ could also be solution).
Then the square root is defined for $2m+1\ge 0$ that is $m\ge -\frac 12$ and $x(-\frac 12)=0$
Since $x(m)\nearrow$ then the range for $x$ is $[0,+\infty)$.
Note: you seem confused in your last line (moreover $m=0$ implies $x=1$ not $\sqrt{3}$).
Another thing you forgot is when simplifying by $x^2$ in the equation $x^2(1-x^2)=x^2(-2m)$.
To be able to do that we need $x\neq 0$.
So $x=0$ is also solution, but reporting in original equation, this is true only for $m=0$ too. Anyway, $x=0$ can be reached also via $m=-\frac 12$.
A: When you omit x^2 on both sides you should assume that x is not 0 and at the end you should say that when $2m +1 >= 0 $ the square root would be true hence $ m>=-1/2$, also $$ x =+\sqrt{(2m + 1)} $$ or $$ x = -\sqrt{ (2m+1)}.$$
But the positive one is only true because $x>=0$.
Also when $ x = 0 $ , $m$ has to be zero as well .
A: $$ \sqrt{ m+\sqrt{x^{2}+m^{2}}}=x $$
$$ m+\sqrt{x^{2}+m^{2}}=x^{2} $$
$$ \sqrt{x^{2}+m^{2}}=x^{2}-m $$
$$ x^{2}+m^{2}=x^{4}+m^{2}-2mx^{2} $$
$$ x^{2}=x^{4}-2mx^{2} $$
$$ x^{2}(1+2m)=x^{4} $$
$$x=\pm\sqrt{1+2m}$$
What is the domain of $x$? What is the domain of $m$, this information determines if the solution is real, complex, valid or non-existent.
