<s>[INST] Alternate methods to solve this equation The question is to find the roots of the equation:

$2(x^2+ \frac{1}{x^2}) - 3(x+ \frac{1}{x}) - 1 =0$

My attempt:
$2(x^2+ \frac{1}{x^2} + 2) - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
⇒$2(x+ \frac{1}{x})^2 - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
⇒$2(x+ \frac{1}{x})^2 - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
Let $(x+ \frac{1}{x})$ be $'a'$
Therefore the given equation becomes,
$2a^2 -3a - 5 = 0$
⇒$2a^2 -a(5-2) - 5 = 0$
⇒$2a^2 -5a + 2a - 5 = 0$
⇒$a(2a - 5) + 1(2a - 5) = 0$
⇒$(2a - 5)(a + 1) = 0$
⇒$(2a - 5) = 0$  or  $(a + 1) = 0$
⇒$a = 5/2$  or  $a = -1$

When $a = 5/2$
⇒$(x+ \frac{1}{x}) = 5/2$
⇒$2(x^2 + 1) = 5x$
⇒$2x^2 - 5x  + 2 = 0$
⇒$2x^2 -x(4 + 1) + 2 = 0$
⇒$2x^2 - 4x - x + 2 = 0$
⇒$2x(x - 2) - 1(x - 2) = 0$
⇒$(x - 2)(2x - 1) = 0$
⇒$(x - 2) = 0$ or $(2x - 1) = 0$
⇒$x = 2$ or  $x = 1/2$

When $a = -1$
$(x+ \frac{1}{x}) = -1$
⇒$x^2 + 1 = -x$
⇒ $x^2 + x + 1 = 0$
⇒$x = \frac {-1 \pm \sqrt{1^2 - 4.1.1}}{2.1}$
⇒$x = \frac {-1 + \sqrt{1^2 - 4.1.1}}{2.1}$ or $\frac {-1 - \sqrt{1^2 - 4.1.1}}{2.1}$
⇒$x = \frac {-1 + \sqrt{3}i}{2}$ or $x = \frac {-1 - \sqrt{3}i}{2}$ 

My question is: Is there any elegant technique to solve this solve because i find my method lengthy and cumbersome.
 [/INST] Hint -
$(x+ \frac{1}{x}) = t$
Also,
$(x^2+ \frac{1}{x^2}) = t^2 - 2$
Hope it helps.
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