Quadratic Dilemma

$\begin{aligned} \frac {x^2}{10} + \frac {28}x & = \frac {7x}5 - \frac {40}{x^2} & \small \color{#3D99F6} \text{Multiply both sides by }10 \\ x^2 + \frac {280}x & = 14x - \frac {400}{x^2} & \small \color{#3D99F6} \text{Put all terms on the left.} \\ x^2 + \frac {400}{x^2} - 14x + \frac {280}x & = 0 \\ {\color{#3D99F6} x^2 - 40 + \frac {400}{x^2}} + {\color{#D61F06}40} - 14 \left( x - \frac {20}x\right) & = 0 \\ {\color{#3D99F6} \left(x - \frac {20}x \right)^2} - 14 \left( x - \frac {20}x\right) + {\color{#D61F06}40} & = 0 & \small \color{#3D99F6} \text{A quadratic equation of }\left(x - \frac {20}x \right) \end{aligned}$

$\begin{aligned} \implies x - \frac {20}x & = \frac {14 \pm \sqrt{14^2-4(40)}}2 = \begin{cases} 10 \\ 4 \end{cases} \end{aligned}$

$\begin{cases} x^2 - 10 x - 20 & = 0 & \implies x = \begin{cases} 5 + 3\sqrt 5 = x_4 \\ 5 - 3\sqrt 5 = x_2 \end{cases} \\ x^2 - 4 x - 20 & = 0 & \implies x = \begin{cases} 2 + 2\sqrt 6 = x_3 \\ 2 - 2\sqrt 6 = x_1 \end{cases} \end{cases}$

By Vieta's formula , we have $x_1x_3=-20$ and $x_2x_4=-20$ , $\implies x_1x_3+ x_2x_4 = \boxed{-40}$

$\begin{aligned} \frac{x^2}{10}+\frac{28}{x}&=\frac{7x}{5}-\frac{40}{x^2}\\ x^2+\frac{280}{x}&=14x-\frac{400}{x^2}\\ x^2+\frac{280}{x}-14x+\frac{400}{x^2}&=0\\ x^2+\frac{400}{x^2}+49-40+\frac{280}{x}-14x&=9\\ \left(x-\frac{20}{x}-7\right)^2&=9 \end{aligned}$

So $x-\frac{20}{x}-7=3\quad$ or $\quad x-\frac{20}{x}-7=-3$

Case 1:

$\begin{aligned} x-\frac{20}{x}-7&=3\\ x^2-10x-20&=0\\ x=5\pm 3\sqrt{5} \end{aligned}$

Case 2:

$\begin{aligned} x-\frac{20}{x}-7&=-3\\ x^2-4x-20&=0\\ x=2\pm 2\sqrt{6} \end{aligned}$ .

When arranged in the order $x_1<x_2<x_3<x_4$ , the roots are $(x_1,x_2,x_3,x_4)=(2-2\sqrt{6}, 5-3\sqrt{5}, 2+2\sqrt{6}, 5+3\sqrt{5})$ .

Hence,

$\begin{aligned} x_1x_3+x_2x_4&=(2-2\sqrt{6})(2+2\sqrt{6})+(5-3\sqrt{5})(5+3\sqrt{5})\\ &=-20+(-20)\\ &=\boxed{-40} \end{aligned}$