Step-by-step Solution:

1. Analyze the given polynomials. We are given the polynomial $f(x) = x^4 + ax^3 + bx^2 + c$ with real coefficients $a, b, c$. The derivative of $f(x)$ with respect to $x$ is $f'(x) = 4x^3 + 3ax^2 + 2bx$. The second equation given is $4x^3 + 3ax^2 + 2bx = 0$, which is equivalent to $f'(x) = 0$.

2. Use the root of $f'(x)=0$ to find coefficients $a$ and $b$. We are given that $x = i\sqrt{3}$ is a root of $f'(x)=0$. Let's substitute this value into the equation: $f'(i\sqrt{3}) = 4(i\sqrt{3})^3 + 3a(i\sqrt{3})^2 + 2b(i\sqrt{3}) = 0$ Let's simplify the powers of $i\sqrt{3}$:

   • $(i\sqrt{3})^2 = i^2 (\sqrt{3})^2 = -1 \cdot 3 = -3$
   • $(i\sqrt{3})^3 = (i\sqrt{3})^2 \cdot (i\sqrt{3}) = -3 \cdot i\sqrt{3} = -3i\sqrt{3}$

   Substituting these back into the equation: $4(-3i\sqrt{3}) + 3a(-3) + 2b(i\sqrt{3}) = 0$ $-12i\sqrt{3} - 9a + 2bi\sqrt{3} = 0$ Group the real and imaginary parts: $(-9a) + i(2b\sqrt{3} - 12\sqrt{3}) = 0$ For this complex number to be zero, both its real and imaginary parts must be zero.

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

   Alternatively, since the coefficients of $f'(x)$ are real ($4, 3a, 2b$), if a complex number $i\sqrt{3}$ is a root, its conjugate $-i\sqrt{3}$ must also be a root. The equation $f'(x)=0$ can be written as $x(4x^2+3ax+2b)=0$. The roots are $0, i\sqrt{3}, -i\sqrt{3}$. The quadratic factor $4x^2+3ax+2b$ has roots $i\sqrt{3}$ and $-i\sqrt{3}$. Using Vieta's formulas for the quadratic:

   • Sum of roots: $i\sqrt{3} + (-i\sqrt{3}) = 0 = -\frac{3a}{4} \implies a=0$.
   • Product of roots: $(i\sqrt{3})(-i\sqrt{3}) = -i^2(3) = 3 = \frac{2b}{4} = \frac{b}{2} \implies b=6$.

3. Use the condition $f(1)=-9$ to find the coefficient $c$. Now that we have $a=0$ and $b=6$, the polynomial is $f(x) = x^4 + 6x^2 + c$. We are given $f(1) = -9$. Let's substitute $x=1$: $f(1) = (1)^4 + 6(1)^2 + c = -9$ $1 + 6 + c = -9$ $7 + c = -9$ $c = -16$

4. Find the roots of $f(x)=0$. The polynomial is $f(x) = x^4 + 6x^2 - 16$. To find the roots, we solve the equation $f(x)=0$: $x^4 + 6x^2 - 16 = 0$ This is a quadratic equation in terms of $x^2$. Let $y = x^2$. The equation becomes: $y^2 + 6y - 16 = 0$ Factoring the quadratic: $(y+8)(y-2) = 0$ This gives two possible values for $y$: $y=-8$ or $y=2$.

   • Case 1: $x^2 = -8 \implies x = \pm \sqrt{-8} = \pm i\sqrt{8} = \pm 2i\sqrt{2}$.
   • Case 2: $x^2 = 2 \implies x = \pm \sqrt{2}$.

   So, the four roots of $f(x)=0$ are $\alpha_1 = 2i\sqrt{2}$, $\alpha_2 = -2i\sqrt{2}$, $\alpha_3 = \sqrt{2}$, and $\alpha_4 = -\sqrt{2}$.

5. Calculate the required sum. We need to find the value of $\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2$. Let's calculate the squared magnitude for each root:

   • $|\alpha_1|^2 = |2i\sqrt{2}|^2 = (2\sqrt{2})^2 = 4 \cdot 2 = 8$.
   • $|\alpha_2|^2 = |-2i\sqrt{2}|^2 = (2\sqrt{2})^2 = 4 \cdot 2 = 8$.
   • $|\alpha_3|^2 = |\sqrt{2}|^2 = (\sqrt{2})^2 = 2$.
   • $|\alpha_4|^2 = |-\sqrt{2}|^2 = (\sqrt{2})^2 = 2$.

   The sum is: $8 + 8 + 2 + 2 = 20$

Thus, the value of $\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2$ is 20.