How to simplify $\left({{\frac{2207+(2207^2-4)^{1/2}}2}}\right)^{1/8}$ into quadratic form? The 1995 Putnam B4 asks to evaluate some infinite expression and I got this value after some computations:$$\left({{\frac{2207+(2207^2-4)^{1/2}}2}}\right)^{1/8}$$
However I am supposed to write it in form:$$\frac{a+b\sqrt{c}}{d}$$. What do I do?
(The answer is: $\frac{3+\sqrt{5}}{2}$ which is the same as mine if you put in a calculator.)

Let $\,x \gt 1\,$ be the expression in question, then:
$$\require{cancel}
2x^8=2207 + \sqrt{2207^2-4} \\
(2x^8-2207)^2=2207^2-4 \\
4x^{16}-4\cdot 2207 x^8+\cancel{2207^2}=\cancel{2207^2}-4 \\
x^{16}-2207 x^8 + 1 = 0 \\
x^8 + \frac{1}{x^8}=2207 \\
\left(x^4+\frac{1}{x^4}\right)^2 = 2209 \\
x^4+\frac{1}{x^4} = 47 \\
\left(x^2+\frac{1}{x^2}\right)^2 = 49 \\
x^2+\frac{1}{x^2} = 7 \\
\left(x+\frac{1}{x}\right)^2 = 9 \\
x+\frac{1}{x} = 3 \\
x^2-3x+1=0
$$
