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

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 $(x^2+7x+1)$.
