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

Question

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.

Answer 1

So you've got this: $$\sqrt{(x-a)(x-b)}+\sqrt{(x-b)(x-c)}=0$$ And the $\sqrt{x}$ is always non-negative. So you are adding $2$ non-negative numbers together. When can that sum be $0$? When both of the terms are $0$. So you should have both $\sqrt{(x-a)(x-b)}=0$ and $\sqrt{(x-b)(x-c)}=0$. But $\sqrt{x}=0$ only if $x=0$, so you should have $(x-a)(x-b)=0$ and $(x-b)(x-c)=0$. From this, $x=b$ is a solution, and if $a=c$, then $x=a=c$ is also a solution.

Answer 2

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.

Answer 3

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 $x=b$)