Let $f(x)=x^4+a x^3+b x^2+c$ be a polynomial with real coefficients such that $f(1)=-9$. Suppose that $i…
Let $f(x)=x^4+a x^3+b x^2+c$ be a polynomial with real coefficients such that $f(1)=-9$. Suppose that $i \sqrt{3}$ is a root of the equation $4 x^3+3 a x^2+2 b x=0$, where $i=\sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$, and $\alpha_4$ are all the roots of the equation $f(x)=0$, then $\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2$ is equal to ________.
Solution
$f(1)=1+a+b+c=-9 \quad \Rightarrow \quad a+b+c=-10.....(1)$
$4 x^3+3 a x^2+2 b x=0$ roots are $\sqrt{3} i,-\sqrt{3} i, 0$
$\Rightarrow4 x^2+3 a x+2 b=0 < ^{\sqrt{3i}}_{-{\sqrt{3i}}}$
$\Rightarrow \quad \mathrm{a}=0 ~\&~ \frac{2 \mathrm{~b}}{4}=(\sqrt{3} \mathrm{i})(-\sqrt{3} \mathrm{i})$
$\mathrm{b}=6$ use $\mathrm{a}, \mathrm{b}$ in (1) $\Rightarrow \mathrm{c}=-16$
$\begin{array}{ll}\Rightarrow & \mathrm{f}(\mathrm{x})=\mathrm{x}^4+6 \mathrm{x}^2-16=0 \\ & \left(\mathrm{x}^2+8\right)\left(\mathrm{x}^2-2\right)=0 \\ \Rightarrow \quad & \mathrm{x}= \pm \sqrt{8} \mathrm{i}, \pm \sqrt{2}\end{array}$
$\Rightarrow \quad\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2=20$