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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. . . . . .
$10$
$20$
$30$
$40$
Solution
$f(1)=1+a+b+c=-9 \quad \Rightarrow \quad a+b+c=-10$ $. . . . (1)$
$4 x^3+3 a x^2+2 b x=0 \text { roots are } \sqrt{3} i,-\sqrt{3} i, 0$
$\Rightarrow \quad 4 x ^2+3 ax +2 b =0 < {l} \sqrt{3} i$
$-\sqrt{3} i$
$\Rightarrow \quad a =0 \& \frac{2 b }{4}=(\sqrt{3} i )(-\sqrt{3} i )$
$b =6 \text { use } a , b \text { in (1) } \Rightarrow c =-16$
$\Rightarrow \quad f(x)=x^4+6 x^2-16=0$
$\left(x^2+8\right)\left(x^2-2\right)=0$
$\Rightarrow \quad x = \pm \sqrt{8} i , \pm \sqrt{2} \quad \Rightarrow \quad\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2=20$
