<s>[INST] $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$ In MAO 1991,

Find $2x+5$ if $x$ satisfies $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$

My attempt,
I squared the the equation then I got $144x^2+1648x+4480=144x^2-1632x+4624$, which results $x=-9$, and $2x+5=-13$.
I want to ask is there another way to solve this question as my method is very tedious. Thanks in advance.
 [/INST] An $x$ that solves your equation also solves the system
\begin{align} 
40&&-9x&-a^2&=0\\
7&&-x&-b^2&=0\\
&&-x&-c^2&=0\\
&&a-2b&-c&=0\\
\end{align}
If we subtract the second and the third equation we get 
$$-b^2+c^2+7=0.$$
It turns out that 
$$-9(-b^2+c^2+7)=-(3b-4c)(-3b-4c)-(-7c^2+63).$$
Therefor one solution is $c^2-9=0$ and $3b-4c=0$. Both solutions to $c$ work. If we pick $c=3$ we get $x=-9$ from the third equation and $b=4$ which makes $x=-9$ consistent with the second equation.
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