Infinite Roots

Let $x=\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-...}}}}$

Squaring both sides,

$x^2=7+\sqrt{7-\sqrt{7+\sqrt{7-...}}} \implies x^2-7=\sqrt{7-\sqrt{7+\sqrt{7-...}}}$

Once again, by squaring both sides,

$\left(x^2-7\right)^2=7-\sqrt{7+\sqrt{7-...}} \implies \left(x^2-7\right)^2-7=-\left(\sqrt{7+\sqrt{7-...}}\right) \implies \left(x^2-7\right)^2-7=-x$

$\implies x^4-14x^2+x+42=0$

Using the rational root theorem, we find that $x=3$ is a root of the polynomial as well as in the options. Hence, $x=3 \implies \boxed{\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-...}}}}=3}$ .

For those who are wondering about the three other roots: After using the rational root theorem rigorously, I noticed that $x=3$ and $x=-2$ are two of the integral roots of this bi-quadratic function. For obtaining the other two roots, I divided $(x^4-14x^2+x+42)$ by $(x-3)(x+2)$ and got $(x^2+x-7)$ as the quotient and no remainder. Now using the quadratic formula, we have $x=\frac{1}{2} (-1 \pm \sqrt{29})$ . Hence, the three other roots of this bi- quadratic functions are $\boxed{-2, \frac{1}{2} (-1 + \sqrt{29}), \frac{1}{2} (-1 - \sqrt{29})}$ .