Finding $a$ and $b$ for the given polynomial Question:

All the roots of $x^4 – 12x^3 + ax^2 + bx + 81 = 0$ are non-negative. The ordered pair $(a, b)$ can be?
Options: 
A) $(9,36)$ B) $(27,-108)$C) $(54,-108)$ D) $(36,108)$

Here. I eliminated option (A) and (D) as $b$ must be negative. But I was unable to calculate the value of $a$ as it required solving quite complex equations which were beyond my scope. Following were the equations:

$\alpha +\beta +\gamma +\delta = 12$ 
$\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = a$ 
$\alpha \beta \gamma +\alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = -b$ 
$\alpha \beta \gamma \delta = 81$ 

This must not be the way to go about finding $a$ but I can't think of anything else. Please help.
Answer:

 Option (C)

 A: $x^4 – 12x^3 + ax^2 + bx + 81 = 0$
Let $\alpha, \beta ,\gamma ,\delta$ are four non negative roots.
Then

*

*$\alpha +\beta +\gamma +\delta = 12$


*$\alpha \beta \gamma \delta = 81$
Then $\frac{\alpha +\beta +\gamma +\delta}{4}=3=(81)^{\frac{1}{4}}=(\alpha \beta \gamma \delta) ^{\frac{1}{4}}$
Hence Arithmetic mean of $\alpha ,\beta, \gamma ,\delta $ is equal to geometric mean (A.M=G.M) .
Hence $\alpha= \beta= \gamma =\delta $.
From $1$ , $\alpha +\alpha+\alpha+\alpha=12$ implies $\alpha=3$
Hence $3$ is root of multiplicity $4$.
Now from the remaining two relations between roots and coefficients, we have
$\begin{align}a&=\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta\\&=6\cdot 3^2 \end{align}$
$\implies a=54$

$\begin{align}-b&=\alpha \beta \gamma +\alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta\\& =4\cdot 3^3\end{align}$
$\implies b=-108$
A: By the Descartes’ rule of signs, $b$ must be negative and $a$ positive.
Notice that the  constant term of the equation is a perfect square. This is our motivation for the goal to convert the expression to the form (if possible) $$(x^2+cx+9)^2=0.$$ This will give us $x^2+cx+9=0$. Again, by the Descartes’ rule of signs, $c$ must be negative.
Thus, comparing coefficients between $$x^4+2cx^3+(c^2+18)x^2+18cx+81=0$$ and $$x^4 – 12x^3 + ax^2 + bx + 81 = 0,$$ we get $$2c=-12\implies c=-6$$ so that $$a=54, b=-108.$$
