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!
Best Answer
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.