The touch of Madness

We note that $\large (x^7+ \frac{1}{x^7})(x^2+\frac{1}{x^2})=(x^9+\frac{1}{x^9})+(x^5+\frac{1}{x^5})$ =LHS

Using this and the given equation we get

$\large (x^7+ \frac{1}{x^7})(x^2+\frac{1}{x^2})=7(x^7+ \frac{1}{x^7})$

$\large(x^2+\frac{1}{x^2})=7$

$\large (x+\frac{1}{x})^2 -2=7$

$\large (x+\frac{1}{x})^2=9$

$\large (x+\frac{1}{x})=\boxed{3}$ as only positive value is required

Let $S_n=x^n+\frac 1 {x^n}$

Note first that $S_{n+1} = \left(x + \dfrac{1}{x}\right)S_{n} - S_{n-1} = S_{1}S_{n} - S_{n-1},$ (A).

Thus $S_{9} = S_{1}S_{8} - S_{7}$ and $S_{7} = S_{1}S_{6} - S_{5},$ and so

$S_{5} + S_{9} = S_{1}(S_{6} + S_{8}) - 2S_{7}.$

Since we are given that $S_{5} + S_{9} = 7S_{7}$ we see that

$S_{1}(S_{6} + S_{8}) = 9S_{7},$ (B).

But from equation (A) we have that

$S_{8} = S_{1}S_{7} - S_{6} \Longrightarrow S_{6} + S_{8} = S_{1}S_{7},$

which combined with (B) give us that $(S_{1})^{2}S_{7} = 9S_{7} \Longrightarrow (S_{1})^{2} = 9,$

since clearly $S_{7} \ne 0.$ The desired positive value of $S_{1}$ is thus $\boxed{3}.$

x=cos∆+isin∆ substitute and simplify you'll get cos∆ as 3/2