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

Question

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

Answer 1

Let $x+\dfrac1x=y$

$y^2=?$

Solve the quadratic equation in $y$

Answer 2

$$\begin{array}{rcl} 4x^2 + 12x + \dfrac{12}{x} + \dfrac{4}{x^2} &=& 47 \\ 4\left(x^2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 47 \\ 4\left(x^2 + 2 + \dfrac1{x^2}\right) + 12\left(x + \dfrac1x\right) &=& 55 \\ 4\left(x + \dfrac1x\right)^2 + 12\left(x + \dfrac1x\right) &=& 55 \\ 4\left(x + \dfrac1x\right)^2 + 12\left(x + \dfrac1x\right) - 55 &=& 0 \\ \left(x + \dfrac1x\right) &=& -5.5 ~\text{or}~ 2.5 \\ \end{array}$$

Answer 3

\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}