Solve the equation $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$$
$x=5$ is the only real solution
We can note that $x^2-9x+24>0$ and $6x^2-59x+149>0$ for all $x$. The first thing I decided to try: as $$|5-x|=\begin{cases}5-x,x\le5\\x-5,x>5\end{cases},$$ we can look at two different cases based on the sign of $5-x$: let's $x>5$. Then the equation becomes $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=x-5\\\sqrt{x^2-9x+24}=x-5+\sqrt{6x^2-59x+149}\\x^2-9x+24=(x-5)^2+6x^2-59x+149+2(x-5)\sqrt{6x^2-59x+149}\\30x-3x^2-7=(x-5)\sqrt{6x^2-59x+149}$$ We have to raise both sides to the power of 2 again, so I decided to stop here.
Something else we can try is to raise both sides of the initial equation to the power of 2, as $|5-x|^2=(5-x)^2,$ so we have $$7x^2-68x+173-2\sqrt{(x^2-9x+24)(6x^2-59x+149)}=x^2-10x+25\\3x^2-29x+72=\sqrt{(x^2-9x+24)(6x^2-59x+149)}$$ I think it's obvious I am missing something.
Best Answer
Since, $|5-x|\geqslant 0$, then you have:
$$\begin{align}&\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}\geqslant 0\\ \iff &x^2-9x+24\geqslant 6x^2-59x+149\geqslant 0\\ \iff &-5(x-5)^2\geqslant 0\\ \iff &x=5. \end{align}$$
This implies that, if there's a solution, then it is $5$. Indeed, $x=5$ is a solution.
Remember that, sometimes it is easier to find the domain of the equation than to solve the equation itself. That's exactly what we're doing in this equation.