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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.

  • 0
    This is better than knowing the general expression for the roots of degree $4$ polynomial. – 2017-02-15
  • 0
    I quite like your approach. You have some steps in your factoring that I think could be omitted. Anyway, I might have multiplied through by $x^2,$ creating $2x^4 - 3x^3 - x^2 - 3x + 2 = 0$ Rational root theorem says that rational roots if they exists must be in$\{1,-1,2,-2,\frac 12, -\frac 12\}$ – 2017-02-15

1 Answers 1

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Hint -

$(x+ \frac{1}{x}) = t$

Also,

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

Hope it helps.