Solve the equation $\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$ 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.
 A: 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.
A: Observe that
\begin{align*}
x^2-9x+24&=(x-5)^2+(x-5)+4\\
6x^2-59x+149&=6(x-5)^2+(x-5)+4
\end{align*}
Say $(x-5)^2+(x-5)+4=\alpha$ (note $\alpha \geq 0$), then we have
$$\sqrt{\alpha}-\sqrt{5(x-5)^2+\alpha}=|5-x|$$
The expression on the left is $\leq 0$ and that on the right is $\geq 0$. For the equality to hold, both have to be $0$.
Thus $x=5$ is the only solution.
A: Well we must have the following conditions.
$x^2 -9x +24 \ge 0$ (other wise the square root doesn't exist)
$6x^2 - 59x + 149 \ge 0$.  (ditto)
$\sqrt{x^2 - 9x + 24} \ge \sqrt{6x^2 - 59x + 149}$ (because their difference is $|5-x|$ which is $\ge 0$)
$x^2 - 9x + 24 \ge 6x^2 - 59x + 149$ (because $0 \le a \le b \implies a^2 \le b^2$)
So $5x^2 - 50x + 125 \le 0$.  We can factor to get
$5(x-5)^2 \le 0$. and we'll.. that actually throws me throw a loop... I was only trying to find a valid range that $x$ can be in but this is ... big.
$(x-5)^2$ is a perfect square and so $(x-5)^2 \ge 0$ so the only way we can have $5(x-5)^2 \le 0$ is if $(x-5)^2 = 0$ and $x = 5$.
So if there is any answer the only possible answer is $x=5$.
But that doesn't mean that there is any answer.  After all anyone can write down anything.  That doesn't mean it has an answer.
So we have to check that if $x=5$ the equation holds.
If $x =5$ then $x^2 -9x + 24 = 25-45 + 24 = 4$ and $\sqrt{x^2 -9x+24} = 2$.
And $6x^2 - 59x + 149= 150 -295 + 149= 4$ and $\sqrt{6x^2 - 59x + 149} = 2$.
And $|5-x| = |5-5|=0$ and $\sqrt{x^2 -9x+24} - \sqrt{6x^2 - 59x + 149}=2-2=0=|5-x|$ and thus $x = 5$ is a solution.
