If $\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then
a) x=3
b) x= $\displaystyle \infty$
c) x=6
d) x=9
Please tell the answer with complete explaination
Let: $\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$
Now, note that if you square it and move +6 to the LHS: $\displaystyle x^2 = 6 + \sqrt{6+\sqrt{6+ \cdots}} \ \ \Leftrightarrow \ \ x^2 - 6 = \sqrt{6+\sqrt{6+ \cdots}}$
But the RHS is $\displaystyle x$ again!
Thus, we see that $\displaystyle x$ is the positive solution to the quadratic: $\displaystyle x^2 - x - 6 = 0$
which is simply 3.