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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 ?

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

2 Answers 2

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Hint: set $$t=x+\frac{6}{x}$$ and you will get a quadratic equation in $t$

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$t^2+t=30$ iff $t=5$ or $t=-6$, and $$ x+\frac{6}{x}\in\{-6,5\} $$ iff $\color{red}{x\in\{2,3,-3-\sqrt{3},-3+\sqrt{3}\}}$.

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