Solve for $x$ given $4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$ 
$$4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$$ Solve for $x$.

I tried to put every single monomial under $x^2$ denominator, but that did not get me to anything I could solve for.
I appreciate any help.
 A: Let $x+\dfrac1x=y$
$y^2=?$
Solve the quadratic equation in $y$
A: $$\begin{array}{rcl}
4x^2 + 12x + \dfrac{12}{x} + \dfrac{4}{x^2} &=& 47 \\
4\left(x^2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 47 \\
4\left(x^2 + 2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 55 \\
4\left(x + \dfrac1x\right)^2 + 12\left(x + \dfrac1x\right) &=& 55 \\
4\left(x + \dfrac1x\right)^2 + 12\left(x + \dfrac1x\right) - 55 &=& 0 \\
\left(x + \dfrac1x\right) &=& -5.5 ~\text{or}~ 2.5 \\
\end{array}$$
A: \begin{equation}
 4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47
\end{equation}
Multiply by $x^2$, i.e.
\begin{equation}
 4x^4 + 12x^3 -47x^2  +  12x + 4= 0
\end{equation}
Notice that for $x = 2$ and $x = \frac{1}{2}$, we get zero. So 
\begin{equation}
 4(x-2)(x-\frac{1}{2})(x - a)(x-b)
 =
 4x^4 + 12x^3 -47x^2  +  12x + 4
\end{equation}
Expand the left hand side, i.e.
\begin{equation}
 4(x^2 - \frac{5}{2}x + 1)(x^2 - (a+b)x + ab) 
\end{equation}
That is 
\begin{equation}
 (4x^2 - 10x + 4)(x^2 - (a+b)x + ab) 
\end{equation}
Expand now
\begin{equation}
 4x^4 -4(a+b)x^3 + 4abx^2
 -10x^3 + 10(a+b)x^2 -10abx
 +4x^2 -4(a+b)x +4ab
\end{equation}
That is
\begin{equation}
 4x^4 + (-10 -4(a+b))x^3 + (4ab+10(a+b)+4)x^2 +(-10ab
  -4(a+b))x +4ab
\end{equation}
Equate it to the right hand side of the third equation and we get by identification 
\begin{align}
 -10 -4(a+b) &= 12 \\ 
 4ab+10(a+b)+4 &= -47 \\
 -10ab-4(a+b) &= 12  \\
 4ab &= 4
\end{align}
So
\begin{equation}
 ab = 1
\end{equation}
and hence 
\begin{equation}
 (a+b) = -\frac{12 + 10(1)}{4} = -\frac{11}{2}
\end{equation}
Now let's find $a,b$
\begin{align}
 a+b &= -\frac{11}{2}\\
 ab &= 1
\end{align}
So, $b = -a -\frac{11}{2}$ and hence $ab = a(-a -\frac{11}{2}) = 1$ which gives a quadratic system 
\begin{equation}
 a^2 + \frac{11}{2}a +1 = 0
\end{equation}
Giving two roots
\begin{align}
 a_1 &= \frac{1}{4}(-11 - \sqrt{105})\\
 a_2 &= \frac{1}{4}(-11 + \sqrt{105}) 
\end{align}
It turns out that 
\begin{equation}
 b_1= \frac{1}{a_1} = a_2 
\end{equation}
and
\begin{equation}
 b_2= \frac{1}{a_2} = a_1
\end{equation}
So the roots of the initial equation are 
\begin{align}
 x_1 &= 2\\
 x_2 &= \frac{1}{2}\\
 x_3 &= \frac{1}{4}(-11 - \sqrt{105})\\
 x_4 &= \frac{1}{4}(-11 +\sqrt{105})
\end{align}
