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If $x$ satisfies $x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$.

This equation becomes $x^4+3x^3-26x^2+3x+1=0$ which has four solutions. One of the solutions is $2+\sqrt3$ which has the form $a+\sqrt b$. So, $a+b=2+3=5$. How does this look?

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  • $\begingroup$ Completely fine. $\endgroup$ – Parcly Taxel Aug 27 '17 at 14:38
  • $\begingroup$ How did you find that solution? $\endgroup$ – José Carlos Santos Aug 27 '17 at 14:45
  • $\begingroup$ I used Wolfram Alpha. $\endgroup$ – ddswsd Aug 27 '17 at 14:46
  • $\begingroup$ I'm sure there is a clever way to find it algebraically. $\endgroup$ – ddswsd Aug 27 '17 at 14:47
  • $\begingroup$ @ddswsd I am quite sure that whoever invented this problem did not have that in mind. $\endgroup$ – José Carlos Santos Aug 27 '17 at 14:50
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Let $t = x + \frac{1}{x}$, then the equation is $$t^2 - 2 + 3t = 26,$$ or $$t^2 + 3t -28 = 0.$$

Then, we have $t = 4$ or $t = -7$.

If $t=4$, then $x + \frac{1}{x} = 4$, then $x = 2 \pm \sqrt{3}$.

If $t=-7$, then $x + \frac{1}{x} = -7$, then $x = \frac{1}{2}(-7\pm \sqrt{45})$.

Because $a$ and $b$ are positive integers, then $x = 2 + \sqrt{3}$. And yes, $a+b=5$.

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We can factor this equation as follows: $$x^4+3x^3-26x^2+3x+1 = (x^2-4x+1) (x^2+7x+1).$$


One can check that $2\pm \sqrt3$ are solutions to the $(x^2-4x+1)$,
alse we can check that $\dfrac{-7\pm 3\sqrt5}{2}$ are solutions to the $(x^2+7x+1)$.

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