The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6 $, consists of: 
The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6 $, consists of:


$A)$ Two rational numbers. $B)$ An irrational number. $C)$ Complex number. $D)$ None.

How to solve the above question using proper method?
I got the values by substitution. I just considered the values after evaluation of the radical to be $(5,1)$ ,$(3,3)$ and $(4,2)$.
This leads me to nice and whole values of $x = 6$. I can't find any other solutions.
I tried squaring, but the solution is getting really nasty at a point. I wonder if there could be any more methods to solve this problem. Any help or suggestion would be appreciated. Thanks!
 A: Note that\begin{align}\sqrt{4x+1}+\sqrt{7-x}=6&\implies4x+1+7-x+2\sqrt{4x+1}\sqrt{7-x}=36\\&\iff2\sqrt{4x+1}\sqrt{7-x}=28-3x\\&\implies4(4x+1)(7-x)=(28-3x)^2\\&\iff25 x^2-276 x+756=0\\&\iff x=6\text{ or }x=\frac{126}{25}.\end{align}And you can easily check that $6$ and $\frac{126}{25}$ are actually roots of the original equation.
A: WLOG $\sqrt{4x+1}=6\cos^2t$  and $\sqrt{7-x}=6\sin^2t$
$\implies4x+1=36\cos^4t,7-x=36\sin^4t$
$\implies36\cos^4t+144\sin^4t=\cdots=29$
$\implies36\cos^4t+144(1-\cos^2t)^2=\cdots=29$
$\iff180\cos^4t-288\cos^2t+115=0$
$\implies\cos^2t=\dfrac56,\dfrac{23}{30}$
$\implies x=\dfrac{36(\cos^2t)^2-1}4=?$
A: Here's is a variation of Jose's answer, which appeared as I was typing this up:
$$\sqrt{4x+1} + \sqrt{7-x} = 6 \\
\iff \sqrt{4x+1} = 6 - \sqrt{7-x} $$
$$\implies 4x+1 = 36+(7-x) - 12\sqrt{7-x}\tag{1} \\
\iff 5x-42=12\sqrt{7-x} $$
$$\implies 25x^2+1764-420x=144(7-x) \tag{2}\\
\iff 25x^2-276x+756=0 \\
\iff x=6\text{ or }\frac{126}{25}$$
The steps labelled (1) & (2) aren't invertible and might have created extraneous solutions; a quick check verifies that both $6$ and $\frac{126}{25}$ indeed satisfy the original equation.
