Question

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 _____.

Answer 1

Correct answer is : 20

\(\because f(1)=-9 \Rightarrow 1+a+b+c=-9\) ...(1)

\(4 x^{3}+3 a x^{2}+2 b x=0\)

\(\Rightarrow x=0, \quad 4 x^{2}+3 a x+2 b=0\) ...(2)

\(\Rightarrow \sqrt{3} i\) and \(-\sqrt{3} i\) are roots of (2)

\(\Rightarrow \sqrt{3} i-\sqrt{3} i=\frac{-3 a}{4}, \sqrt{3} i(-\sqrt{3} i)=\frac{2 b}{4}\)

\(\Rightarrow a=0, b=6, c=-16\)

\(\Rightarrow f(x)=0 \Rightarrow x^{4}+6 x^{2}-16=0\)

\(\Rightarrow x^{2}=\frac{-6 \pm \sqrt{36+64}}{2}=-3 \pm 5=2,-8\)

\(x=-\sqrt{2},+\sqrt{2},-2 \sqrt{2} i, 2 \sqrt{2} i\)

\(\Rightarrow\left|\alpha_{1}\right|^{2}+\left|\alpha_{2}\right|^{2}+\left|\alpha_{3}\right|^{2}+\left|\alpha_{4}\right|^{2}=20\)