# Solving for $x$ of different powers

I want to solve the following equation for $x$

$$\left(x + \frac{6}{x} \right)^2 + \left( x + \frac{6}{x} \right) = 30$$

I done my working till -

$$x^4 + x^3 - 18x^2 + 6x + 36 = 0$$

From here how do I solve for $x$ when I have any different powers ?

• Why not start with $y=x+{6\over x}$ and solve for $y$ first? Jun 21, 2016 at 16:39
• Or factor the equation you end up with $(x-3)(x-2)(x^2+6x+6)$. Jun 21, 2016 at 16:40
• Note that the equation you started with has a nice structure. The "simplified" quartic, not so much. Jun 21, 2016 at 16:45

Hint: set $$t=x+\frac{6}{x}$$ and you will get a quadratic equation in $t$
$t^2+t=30$ iff $t=5$ or $t=-6$, and $$x+\frac{6}{x}\in\{-6,5\}$$ iff $\color{red}{x\in\{2,3,-3-\sqrt{3},-3+\sqrt{3}\}}$.