JEE Advanced · Mathematics · 2. Quadratic Equations
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 ________.
- A 10
- B 20
- C 30
- D 40
Answer & Solution
Correct Answer
(B) 20
Step-by-step Solution
Detailed explanation
\(f(1)=1+a+b+c=-9 \quad \Rightarrow \quad a~+\) \(b+c=-10.....(1)\)
\(4 x^3+3 a x^2+2 b x=0\) roots are \(\sqrt{3} i,-\sqrt{3} i, 0\)
\(\Rightarrow4 x^2+3 a x+2 b=0 < ^{\sqrt{3i}}_{-{\sqrt{3i}}}\)
\(\Rightarrow \mathrm{a}=0 ~\&~ \frac{2 \mathrm{~b}}{4}=(\sqrt{3} \mathrm{i})(-\sqrt{3} \mathrm{i})\)
\(\mathrm{b}=6\) use \(\mathrm{a}, \mathrm{b}\) in (1) \(\Rightarrow \mathrm{c}=-16\)
\(\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}^4+6 \mathrm{x}^2-16=0 \)
\( \left(\mathrm{x}^2+8\right)\left(\mathrm{x}^2-2\right)=0 \)
\( \Rightarrow \mathrm{x}= \pm \sqrt{8} \mathrm{i}, \pm \sqrt{2}\)
\(\Rightarrow \left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2=20\)
\(4 x^3+3 a x^2+2 b x=0\) roots are \(\sqrt{3} i,-\sqrt{3} i, 0\)
\(\Rightarrow4 x^2+3 a x+2 b=0 < ^{\sqrt{3i}}_{-{\sqrt{3i}}}\)
\(\Rightarrow \mathrm{a}=0 ~\&~ \frac{2 \mathrm{~b}}{4}=(\sqrt{3} \mathrm{i})(-\sqrt{3} \mathrm{i})\)
\(\mathrm{b}=6\) use \(\mathrm{a}, \mathrm{b}\) in (1) \(\Rightarrow \mathrm{c}=-16\)
\(\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}^4+6 \mathrm{x}^2-16=0 \)
\( \left(\mathrm{x}^2+8\right)\left(\mathrm{x}^2-2\right)=0 \)
\( \Rightarrow \mathrm{x}= \pm \sqrt{8} \mathrm{i}, \pm \sqrt{2}\)
\(\Rightarrow \left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2=20\)
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