Solve the equation $\sqrt{x^2-(a+b)x+ab}+\sqrt{x^2-(b+c)x+bc}=0$ where $a$, $b$, and $c$ are real numbers So I realised that you can factor $x^2-(a+b)x+ab$ and $x^2-(b+c)x+bc$ to $(x-a)(x-b)$ and $(x-b)(x-c)$ and then you get  $$\sqrt{(x-a)(x-b)}+\sqrt{(x-b)(x-c)}=0$$ This is where I got stuck. I tried squaring both sides but I couldn't find the solution.
You almost done, $x=b$ is a solution, if $x\not = b$ then you can divide by $\sqrt{(x-b)}$ and you have $\sqrt{(x-a)}+\sqrt{(x-c)} = 0$ now a square is always positive so a sum of two squares is zero if both are zero, hence $x=a$ and $x=c$.

So conclusion: $x=b$ is always a zero, and if $a=c$ then $x=a$ is another solution.
Multiplying both sides with $\sqrt{(x-a)(x-b)}-\sqrt{(x-c)(x-b)}$, we see that necessarily
$$(x-a)(x-b)-(x-c)(x-b)=0,$$
i.e.,
$$ (x-b)(c-a)=0.$$ 
So $x=b$ (which we verify to actually be a solution), or we need $a=c$. 
But with $a=c$, the original equation becomes
$$ 2\sqrt{(x-a)(x-b)}=0,$$
hence $x=a$ (or again $x=b$)