Quartic equation through Golden-ratio

After expanding and rearranging $\dfrac {\sqrt5x^2+1}{\sqrt5x^2-1} + \dfrac {5x-1}{5x+1} = \dfrac 1x + \dfrac {1+\sqrt 5}2$ , we have

$(15\sqrt 5 - 25)x^4 - (5+11\sqrt 5)x^3 + (5+3\sqrt 5)x^2 + (15+\sqrt 5)x + 2 = 0$

By Vieta's formula : $\begin{cases} \begin{aligned} a+b+c+d & = \frac {5+11\sqrt 5}{15\sqrt 5 - 25} \\ ab+ac+ad+bc+bd+cd & = \frac {5+3\sqrt 5}{15\sqrt 5 - 25} \\ abc + abd + acd + bcd & = \frac {15+\sqrt 5}{25 - 15\sqrt 5} \\ abcd & = \frac 2{15\sqrt 5 - 25} \end{aligned} \end{cases}$

Then we have:

$\begin{aligned} \frac 1{a^2} + \frac 1{b^2} + \frac 1{c^2} + \frac 1{d^2} & = \left(\frac 1a+\frac 1b + \frac 1c + \frac 1d \right)^2 - 2 \left(\frac 1{ab}+\frac 1{ac}+\frac 1{ad}+\frac 1{bc}+\frac 1{bd}+\frac 1{cd}\right) \\ & = \left(\frac {abc + abd + acd + bcd}{abcd} \right)^2 - 2 \cdot \frac {ab+ac+ad+bc+bd+cd}{abcd} \\ & = \left(-\frac {15+\sqrt 5}2 \right)^2 - 2 \cdot \frac {5+3\sqrt 5}2 \\ & = (-\varphi - 7)^2 - 2(3\varphi +1) \quad \quad \small \blue{\text{where }\varphi = \frac {1+\sqrt 5}2 \text{ denotes the golden ratio.}} \\ & = \varphi^2 + 8 \varphi + 47 \\ & = 9\varphi + 48 \end{aligned}$

And

$\begin{aligned} \frac 1{a^2} + \frac 1{b^2} + \frac 1{c^2} + \frac 1{d^2} + k \left(\frac 1a+\frac 1b + \frac 1c + \frac 1d \right) & = \sqrt 5 \\ 9\varphi + 48 - (\varphi+7)k & = 2\varphi - 1 \\ (9-k)\varphi + 48 - 7k & = 2\varphi - 1 \\ \implies k = \boxed 7 \end{aligned}$