Comment on $x$, if $x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$. The main question goes like this:
$$x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$$
Comment on $x$.
There are options given as well: 
A) $x$ is an irrational number 
B) $2<x<3$ 
C) $x=3$ 
D) None of these 
My approach:
So I try to square the term first. Pretty obvious, it's not going to get me anywhere, I think. I think $x$ is irrational, but I don't have any solid reason. Please help me out.
 A: Let $a_{n+1}=\sqrt{6+a_n}$, $a_1=\sqrt{6}$.
We'll prove that $a_n<3$.
Since $a_1<3$, it remains to prove that $a_{n+1}<3$ for $a_n<3$.
Indeed, $a_{n+1}=\sqrt{a_n+6}<\sqrt{3+6}=3$.
Id est, $a_n<3$.
By another hand, $a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{(3-a_n)(2+a_n)}{\sqrt{a_n+6}+a_n}>0$,
which says that there is $\lim\limits_{n\rightarrow+\infty}a_n$.
Let $\lim\limits_{n\rightarrow+\infty}a_n=A$.
Hence, $\lim\limits_{n\rightarrow+\infty}a_{n+1}^2=\lim\limits_{n\rightarrow+\infty}a_n+6$ or $A^2=A+6$ or $A=3$, which gives the answer. 
A: Solution C is correct.
Since $x=\sqrt{6+\sqrt{6+\cdots }}$ we have that:
$$ x=\sqrt{6+x}$$
This reduces to the quadratic equation $x^2-x-6=0$ which has solutions $x=-2$ or $x=3$. $x=-2$ is not a solution since $-2\neq \sqrt{6-2}=\sqrt{4}=2$. Thus $x=3$.
A: $$x=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$$$$x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=6+x$$$$x^2-x-6=(x+2)(x-3)=0$$$$x=-2~or~3$$
$x=-2$ is obviously a bad answer
Therefore, $x=3$
To sate the appetite for rigor of the commenters below:
Yes, the series does converge. This is because $3<\sqrt{6+y}<y$ if $y>3$ and $y<\sqrt{6+y}<3$ if $-6<y<3$. 
A: An empirical approach:
Starting fom $0$, the first iterates (truncated) are
$$2.449,2.907,2.984,2.997,3.000$$
Hence you can conjecture the value $3$, and easily prove that $3$ is indeed a possible solution as
$$\sqrt{6+3}=3.$$
Unfortunately, this is technically inconclusive as it doesn't rule out the existence of another solution or the possibility of non-convergence.

Now, studying convergence, we can write
$$3-\lambda\epsilon<\sqrt{6+(3+\epsilon)}<3+\lambda\epsilon$$
which simplifies to
$$-6\lambda+\lambda^2\epsilon<1<6\lambda+\lambda^2\epsilon.$$
This shows that sufficiently close to the value $3$, we have linear convergence  with a common ratio approaching $1/6$. This is confirmed by $$\sqrt{6+3.000006}\approx 3.000001$$ with an excellent approximation.
