Let $f(x)=x^4+a{x}^3+b{x}^2+c$ be a polynomial with real coefficients such that f(1) = -9. Suppose that i√3 is a root of the equation $4x^3+3a{x}^2+2bx=0$, where $i=\sqrt{-<!--\; -\; -->1}$. If ${\mathrm{\alpha <!--\; \alpha \; -->}}_1,{\mathrm{\alpha <!--\; \alpha \; -->}}_2,{\mathrm{\alpha <!--\; \alpha \; -->}}_3$ and ${\mathrm{\alpha <!--\; \alpha \; -->}}_4$ are all the roots of the equation $f(x)=0,$then ${\left|{\mathrm{\alpha <!--\; \alpha \; -->}}_1\right|}^2+{\left|{\mathrm{\alpha <!--\; \alpha \; -->}}_2\right|}^2+{\left|{\mathrm{\alpha <!--\; \alpha \; -->}}_3\right|}^2+{\left|{\mathrm{\alpha <!--\; \alpha \; -->}}_4\right|}^2$ is equal to _______.

The correct answer is 20.

Let the given polynomial with real coefficients be:

$f(x)=x^4+a{x}^3+b{x}^2+c$

Differentiating $f\left(x\right)$ with respect to $x$, we get:

$f^{\prime }(x)=4x^3+3a{x}^2+2bx$

We are given that $i\sqrt{3}$ is a root of the equation $4x^3+3a{x}^2+2bx=0$ (which is $f^{\prime }\left(x\right)=0$). Since the coefficients $a$ and $b$ are real, any complex roots must occur in conjugate pairs. Therefore, $-i\sqrt{3}$ is also a root of $f^{\prime }\left(x\right)=0$.

Since the equation $f^{\prime }\left(x\right)=0$ is cubic, let its third root be $r$. We can write:

$4x^3+3a{x}^2+2bx=4(x-r)(x-i\sqrt{3})(x+i\sqrt{3})$

Simplifying the right-hand side:

$4x^3+3a{x}^2+2bx=4(x-r)(x^2+3)$

$4x^3+3a{x}^2+2bx=4x^3-4r{x}^2+12x-12r$

Comparing the coefficients of both sides, we get:
1) The constant term: $-12r=0\Rightarrow r=0$
2) The coefficient of $x^2$: $3a=-4r=0\Rightarrow a=0$
3) The coefficient of $x$: $2b=12���b=6$

Substituting $a=0$ and $b=6$ back into the expression for $f\left(x\right)$, we have:

$f(x)=x^4+6x^2+c$

We are given that $f\left(1\right)=-9$:

$f(1)=1^4+6(1^2)+c=-9$

$7+c=-9\Rightarrow c=-16$

Thus, the polynomial is:

$f(x)=x^4+6x^2-16$

To find the roots of $f\left(x\right)=0$, let $y=x^2$:

$y^2+6y-16=0$

$(y+8)(y-2)=0$

So, the solutions for $y$ are $y=2$ and $y=-8$.
This gives the four roots ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ as follows:
- From $x^2=2$, we have ${\alpha }_1=\sqrt{2}$ and ${\alpha }_2=-\sqrt{2}$.
- From $x^2=-8$, we have ${\alpha }_3=i\sqrt{8}$ and ${\alpha }_4=-i\sqrt{8}$.

Now, we compute the sum of the squares of the magnitudes of these roots:

${|{\alpha }_1}^2+{|{\alpha }_2}^2+{|{\alpha }_3}^2+{|{\alpha }_4}^2={\left|\sqrt{2}\right|}^2+{|-\sqrt{2}|}^2+{\left|i\sqrt{8}\right|}^2+{|-i\sqrt{8}|}^2$

$=2+2+8+8=20$