Polynomials and Complex Roots
DAILY_CHALLENGE
Grade None
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 ___.
Step-by-Step Solution
Key Concept: iā3 root of f'(x)/x forces a=0 and b=6; then use f(1)=-9 to find c
$4x^3+3ax^2+2bx=x(4x^2+3ax+2b)=f'(x)$.
$i\sqrt{3}$ is a root of $4x^2+3ax+2b=0$. Since coefficients are real, $-i\sqrt{3}$ is also a root.
Sum of roots: $i\sqrt{3}+(-i\sqrt{3})=0=-\dfrac{3a}{4}\Rightarrow a=0$.
Product of roots: $(i\sqrt{3})(-i\sqrt{3})=3=\dfrac{2b}{4}\Rightarrow b=6$.
$f(x)=x^4+6x^2+c$. $f(1)=1+6+c=-9\Rightarrow c=-16$.
$f(x)=x^4+6x^2-16=(x^2+8)(x^2-2)$.
Roots: $\pm2i\sqrt{2}$ (with $|\cdot|^2=8$ each) and $\pm\sqrt{2}$ (with $|\cdot|^2=2$ each).
$\sum|\alpha_i|^2=8+8+2+2=20$.
Correct Answer: 20