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 √(3) is a root of the equation 4 x 3 +3 a x 2 +2 b x=0 , where i=√(-1) . If α 1 , α 2 , α 3 , and α 4 are all the roots of the equation f(x)=0 , then α 1 2 + α 2 2 + α 3 2 + α 4 2 is equal to

Mathematics · Complex Numbers

JEE Advanced 2024 — Paper 1 — Question 9

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 $\qquad$

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

$\quad 4 x^{2}+3 a x+2 b=4\left(x^{2}+3\right)+ (3ax + (2b-12))$

$\Rightarrow a=0 ; b=6$ $f(x)=x^{4}+6 x^{2}+c$

$\mathrm{f}(1)=7+\mathrm{c}=-9 \Rightarrow \mathrm{c}=-16$

$\Rightarrow f(x)=x^{4}+6 x^{2}-16=\left(x^{2}+8\right)\left(x^{2}-2\right)$

$\Rightarrow x= \pm 2 \sqrt{2 i}, \pm \sqrt{2}$ $\Rightarrow\left|\alpha_{1}\right|^{2}+\left|\alpha_{2}\right|^{2}+\left|\alpha_{3}\right|^{2}+\left|\alpha_{4}\right|^{2}=20$ .

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Complex Numbers
Topic: Demoivre's Theorem and Roots of Unity