JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Quadratic Equations and Inequalities

From $f(1)=-9$: $1+a+b+c=-9$, so $a+b+c=-10$ ... (1)

Since $i\sqrt{3}$ is a root of $4x^3+3a{x}^2+2bx=0$, and the polynomial has real coefficients, $-i\sqrt{3}$ is also a root. The third root is 0.

For the polynomial $4x^3+3a{x}^2+2bx=x(4x^2+3ax+2b)$, the roots of $4x^2+3ax+2b=0$ are $\pm i\sqrt{3}$.

Sum of roots: $i\sqrt{3}+(-i\sqrt{3})=0=-\frac{3a}{4}$, so $a=0$.

Product of roots: $(i\sqrt{3})(-i\sqrt{3})=3=\frac{2b}{4}$, so $b=6$.

From (1): $0+6+c=-10$, so $c=-16$.

Thus $f(x)=x^4+6x^2-16=(x^2+8)(x^2-2)=0$.

Roots: $x=\pm 2i\sqrt{2},\pm \sqrt{2}$.