Sum of Absolute Squares of Polynomial Roots with Real Coefficients

To find the value of $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2$, we start by using the given conditions to determine the real coefficients $a$, $b$, and $c$ of the polynomial $f(x) = x^4 + ax^3 + bx^2 + c$

Step 1: Evaluate $f(1) = -9$ Substitute $x = 1$ into $f(x)$: $f(1) = 1^4 + a(1)^3 + b(1)^2 + c = -9$ $1 + a + b + c = -9 \implies a + b + c = -10 \quad \text{--- (1)}$

Step 2: Use the root condition for the auxiliary equation We are given that $i\sqrt{3}$ is a root of the equation $4x^3 + 3ax^2 + 2bx = 0$, where $i = \sqrt{-1}$. Substituting $x = i\sqrt{3}$: $4(i\sqrt{3})^3 + 3a(i\sqrt{3})^2 + 2b(i\sqrt{3}) = 0$

Using $i^2 = -1$ and $i^3 = -i$: $(i\sqrt{3})^2 = -3$ $(i\sqrt{3})^3 = -3\sqrt{3}i$

Substituting these values back into the equation: $4(-3\sqrt{3}i) + 3a(-3) + 2b(\sqrt{3}i) = 0$ $-12\sqrt{3}i - 9a + 2b\sqrt{3}i = 0$ $-9a + i\sqrt{3}(2b - 12) = 0$

Since $a$ and $b$ are real numbers, the real and imaginary parts must independently equal zero:

1. Real part: $-9a = 0 \implies a = 0$
2. Imaginary part: $\sqrt{3}(2b - 12) = 0 \implies 2b = 12 \implies b = 6$

Step 3: Determine $c$ and write $f(x)$ Substitute $a = 0$ and $b = 6$ into equation (1): $0 + 6 + c = -10 \implies c = -16$

Thus, the polynomial $f(x)$ is: $f(x) = x^4 + 6x^2 - 16$

Step 4: Find the roots of $f(x) = 0$ Set $f(x) = 0$: $x^4 + 6x^2 - 16 = 0$

Factoring the quadratic in $x^2$: $(x^2 + 8)(x^2 - 2) = 0$

This yields two cases:

1. $x^2 = 2 \implies x = \pm\sqrt{2}$ $\alpha_1 = \sqrt{2}, \quad \alpha_2 = -\sqrt{2}$ $|\alpha_1|^2 = 2, \quad |\alpha_2|^2 = 2$

2. $x^2 = -8 \implies x = \pm 2\sqrt{2}i$ $\alpha_3 = 2\sqrt{2}i, \quad \alpha_4 = -2\sqrt{2}i$ $|\alpha_3|^2 = |2\sqrt{2}i|^2 = 8, \quad |\alpha_4|^2 = |-2\sqrt{2}i|^2 = 8$

Step 5: Calculate the required sum $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 = 2 + 2 + 8 + 8 = 20$