JEE Advanced 2024 Paper 1 · Q09 · Complex Numbers

Question

Let $f(x) = x^4 + ax^3 + bx^2 + c$ be a polynomial with real coefficients such that $f(1) = -9$. Suppose that $i\sqrt{3}$ is a root of the equation $4x^3 + 3ax^2 + 2bx = 0$, 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 $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2$ is equal to ______.

SolveKar AI · Accepted Answer

$f'(x)=4x^3+3ax^2+2bx$. Given $i\sqrt{3}$ is a root: $4(i\sqrt{3})^3+3a(i\sqrt{3})^2+2b(i\sqrt{3})=0\Rightarrow -12i\sqrt{3}-9a+2b\,i\sqrt{3}=0$. Equate real & imaginary: real $-9a=0\Rightarrow a=0$; imaginary $-12+2b=0\Rightarrow b=6$. So $f(x)=x^4+6x^2+c$. Apply $f(1)=1+6+c=-9\Rightarrow c=-16$. $f(x)=x^4+6x^2-16=(x^2-2)(x^2+8)$. Roots: $\pm\sqrt{2},\pm 2i\sqrt{2}$. Sum of $|\alpha_k|^2=2+2+8+8=20$.

JEE Advanced 2024 Paper 1 · Q09 · Complex Numbers

Solution