Question 9

Let $g(x) = 4x^3 + 3ax^2 + 2bx = x(4x^2 + 3ax + 2b)$. One root is $x=0$. The other roots are from $4x^2 + 3ax + 2b = 0$. Since $i\sqrt{3}$ is a root and coefficients are real, $-i\sqrt{3}$ must also be a root. Sum of roots: $i\sqrt{3} + (-i\sqrt{3}) = 0 = -\frac{3a}{4} \implies a = 0$. Product of roots: $(i\sqrt{3})(-i\sqrt{3}) = 3 = \frac{2b}{4} \implies b = 6$. Given $f(1) = -9 \implies 1^4 + a(1)^3 + b(1)^2 + c = -9$. Substituting $a=0, b=6$: $1 + 0 + 6 + c = -9 \implies c = -16$. So $f(x) = x^4 + 6x^2 - 16$. Roots of $f(x) = 0$: Let $x^2 = t \implies t^2 + 6t - 16 = 0 \implies (t+8)(t-2) = 0$. $t = 2 \implies x = \pm\sqrt{2}$. Let $\alpha_1 = \sqrt{2}, \alpha_2 = -\sqrt{2}$. $t = -8 \implies x = \pm i\sqrt{8}$. Let $\alpha_3 = i\sqrt{8}, \alpha_4 = -i\sqrt{8}$. Magnitudes squared: $|\alpha_1|^2 = 2, |\alpha_2|^2 = 2, |\alpha_3|^2 = 8, |\alpha_4|^2 = 8$. Sum = $2 + 2 + 8 + 8 = 20$.