Quadratic EquationHard
Question
If all the roots of the equation $x^{4} - 12x^{3} + ax^{2} + bx + 81 = 0$ where $a,b \in R$ are positive, then
Options
A.$b + 2a = 0$
B.$b + a = - 54$
C.$a = 54$
D.$a = 27$
Solution
$x_{1}x_{2}x_{3}x_{4} = 81$
$$\begin{matrix} & & x_{1} + x_{2} + x_{3} + x_{4} & = 12 \\ \Rightarrow & \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} & & = \left( x_{1}x_{2}x_{3}x_{4} \right)^{1/4} \\ \Rightarrow & x_{1} & & = x_{2} = x_{3} = x_{4} = 3 \\ \Rightarrow & (x - 3)^{4} & & = 0 = x^{4} - 12x^{3} + 54x^{2} - 108x + 81 = 0 \\ a & & & = 54,\text{ }b = - 108 \end{matrix}$$
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