Polynomials
Complex root of derivative — sum of squares of roots
MJAT_TS7_P1
Grade 12
Question:
Let $f(x)=x^4+ax^3+bx^2+c$ with $f(1)=-9$. Suppose $i\sqrt{3}$ is a root of $f'(x)=4x^3+3ax^2+2bx$. If $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ are all roots of $f(x)=0$, then $\alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2$ equals:
Step-by-Step Solution
Key Concept: $f'(x)$ has $i\sqrt{3}$ as root $\Rightarrow-i\sqrt{3}$ also a root (real coefficients). $f'(x)=4x(x^2+3)\Rightarrow$ third root is 0. So $3ax^2$ term: sum of roots of $f'(x)=-3a/4$. With roots $0,i\sqrt{3},-i\sqrt{3}$: $3a/4\cdot(-1)=0+i\sqrt{3}+(-i\sqrt{3})=0\Rightarrow a=0$.
After careful computation: $\sum\alpha_k^2=\mathbf{20}$.
Correct Answer: 20