How to elegantly solve $\left(x^2-\frac1{x^2}\right)^2+2\left(x+\frac1{x}\right)^2 = 2024$?

Question

Find all values of $x$ such that $$\left(x^2-\frac{1}{x^2}\right)^2\:+\:2\,\left(x+\frac{1}{x}\right)^2 \:=\: 2024$$

I arrived at the solutions $\,x=\pm \sqrt{22 \pm \sqrt{483}}$, $\,\pm \sqrt{-23 \pm 4 \sqrt{33}}\,$, by simplifying to a point where there is no more linear term, and then substituting $y=x^2$. It took many steps and was quite messy.

Is there some elegant way to solve it?

Answer 1

I think your answer is a little off -- as it's always complex.

Observe that $(x\pm\tfrac 1x)^2=x^2\pm2+\tfrac 1{x^2}$ so $(x-\tfrac 1x)^2= (x+\tfrac 1x)^2-4$.

Thus, if we set $u=(x+\tfrac 1x)^2$ your equation is $u(u-4)+2u-2024=0$. The solutions are $u=1 \pm \sqrt{2025}=\{46,-44\}$.

Now we have $x+\tfrac 1x=\pm \sqrt{46}$:$$ x=\frac{\sqrt{46} \pm \sqrt{42}}{2},\frac{-\sqrt{46} \pm \sqrt{42}}{2} $$

Answer 2

How lucky to get the equation presented in this form, this allows us to fairly quickly determine all its solutions. Not sure if it's elegant, but surely it is effective.

Two observations to start with:
$\,(1)\,$ If $x$ is a solution, then $x\neq 0,$ and $\,-x\,$ and $\,\raise{.1em}{\pm}\,\dfrac{1}x\,$ are also solutions.
$(2)\,$ The first summand of the equation would yield the terms $x^4$ and $x^{-4}$, having the highest and the lowest degree overall respectively. Transformation into a polynomial would thus result in an 8th order equation, and we'd expect 8 roots in total.

Using the identities $$x^2-\frac{1}{x^2}\:=\: \left(x+\frac{1}{x}\right)\left(x-\frac1x\right) \quad\text{and}\quad\left(x-\frac1x\right)^2 +2 \:=\: \left(x +\frac1x\right)^2 -2$$ the given equation is turned into $$\left(x+\frac{1}{x}\right)^2 \left[\left(x +\frac1x\right)^2 -2\,\right] \;=\; 2024 = 45^2-1 \\[4ex] \quad\implies\quad \left[\frac14\left(x +\frac1x\right)^2 -\frac{46}4\,\right] \left[\frac14\left(x +\frac1x\right)^2 +\frac{44}4\,\right] \;=\; 0$$

Consider the function $x\mapsto\frac12\left(x +\frac1x\right):\mathbb C\backslash\{0\}\to\mathbb C$. For each range value $a\in\mathbb C$ there are two pre-images $x_\pm$, except for $a=\pm 1$. This follows from resolving $$\frac12\left(x +\frac1x\right)\:=\: a \quad\implies\; x_\pm\:=\: a \pm \sqrt{a^2-1} \quad\text{and }\; x_+\cdot x_- =1\,.$$ When restricted to the real domain, then the function maps $\,\mathbb R\backslash\{0\}\twoheadrightarrow(-\infty, -1]\cup [1,\infty)\,$.

Applying this to the factored equation yields $$\begin{align} a\quad & \qquad\text{solutions}\\[1ex] \frac12\sqrt{46} &\qquad \frac12\sqrt{46}\pm\frac12\sqrt{42} \\[1ex] -\frac12\sqrt{46} &\qquad -\frac12\sqrt{46}\pm\frac12\sqrt{42} \\[1ex] i\sqrt{11} &\qquad \left(\sqrt{11}\pm\sqrt{12}\,\right)i\\[1ex] -i\sqrt{11} &\qquad -\left(\sqrt{11}\pm\sqrt{12}\,\right)i \end{align}$$ The non-real solutions are purely imaginary.