Solving $x + \sqrt{x(x-a)} = b + a/2$ This algebraic equation came up in some work I was doing, and I haven't been able to solve it. I know what the solution is, but I'm quite bothered by not knowing the techniques to get there.
$$x + \sqrt{x(x-a)} = b + \frac{a}{2} $$
My attempts:
$$x + \sqrt{x(x-a)} = \frac{x^2 - x(x-a)}{x - \sqrt{x(x-a)}}$$
This led to the same issues in the denominator that I was having previously.
$$(x + \sqrt{x(x-a)})^2= x^2 + x(x-a) + 2x \sqrt{x(x-a)} = \frac{(a + 4b^2)^2}{2}$$
This RHS looks fairly close to the actual solution for $x $, but I don't see how to work with this unruly equation
 A: $x+\sqrt{x(x-a)}=b+\frac a2\iff 2\sqrt{x(x-a)}=2b+a-2x\quad$ we square this
$\require{cancel}\implies 4x(x-a)=\cancel{4x^2}-\cancel{4ax}=4b^2+a^2+\cancel{4x^2}-8bx+4ab-\cancel{4ax}$
$\implies 8bx=4b^2+a^2+4ab=(2b+a)^2$
First if $b=0$ then $a=0$ is forced and the equation reduces to $x+|x|=0$ which gives whole $x\in\mathbb R^-$ solution.
Then for $b\neq 0$ we have $\displaystyle{x=\frac{(2b+a)^2}{8b}}$.
But as I stated in the comment, finding this does not end the resolution of the problem, we have to check for two conditions : 
$\begin{cases}x(x-a)\ge 0 \\[2ex] b+\frac a2-x\ge 0\end{cases}$
So let's calculate them :
$\displaystyle{x(x-a)=\frac{(a+2b)^2(a-2b)^2}{64b^2}\ge 0}\quad$ this is ok.
$\displaystyle{b+\frac a2-x=\frac{4b^2-a^2}{8b}}\quad$  we need $|a|\le |2b|$ for $b>0$ and the opposite when $b<0$.

So to conclude :
  
  
*
  
*If $a=b=0$ then any $x\in\mathbb R^-$ is solution.
  
*If $b>0$ and $|a|\le |2b|$ then $x=\frac{(2b+a)^2}{8b}$
  
*If $b<0$ and $|a|\ge |2b|$ then $x=\frac{(2b+a)^2}{8b}$
  
*In all other cases there are no solutions in $\mathbb R$
  

As I said, this really important to report in the original equation (in this case, it means checking the signs of various stuff). You cannot just state that $x=f(a,b)$ might be solution, you have to verify for which values of $a,b$ this is really true.
A: Well, we have that:
$$x+\sqrt{x\left(x-\text{a}\right)}=\text{b}+\frac{\text{a}}{2}\tag1$$
Subtract $x$ from both sides:
$$\sqrt{x\left(x-\text{a}\right)}=\text{b}+\frac{\text{a}}{2}-x\tag2$$
Square both sides:
$$\left(\sqrt{x\left(x-\text{a}\right)}\right)^2=\left(\text{b}+\frac{\text{a}}{2}-x\right)^2=x\left(x-\text{a}\right)=\frac{\text{a}^2}{4}+\text{a}\text{b}+\text{b}^2-x\left(\text{a}+2\text{b}\right)+x^2\tag3$$
So:
$$x\left(x-\text{a}\right)-\frac{\text{a}^2}{4}-\text{a}\text{b}-\text{b}^2+x\left(\text{a}+2\text{b}\right)-x^2=$$
$$\text{a}^2-4\text{a}\text{b}-4\text{b}^2+8\text{b}x=0\tag4$$
