Simplify the sum of the square of a polynomial having a fourth degree. Today I learn about polynomial. Because I want to improve my knowledge. Thank you for your support and time for sharing information and experience.
From question :
If $a, b, c$ and $d$ are the roots of polynomial $Ax^4+Bx^3+Cx^2+Dx+E$ then find the value of
$a^2+b^2+c^2+d^2$
What I know :

*

*$a+b+c+d=\frac{-B}{A}$

*$ab+ac+ad+bc+bd+cd=\frac{C}{A}$

*$abc+abd+acd+bcd=\frac{-D}{A}$

*$abcd=\frac{E}{A}$
What I try :
$$(a+b+c+d)^2=(a+b+c+d)(a+b+c+d)$$
$$\left(\frac{-B}{A}\right)^2=(a^2+ab+ac+ad)+(ab+b^2+bc+bd)+(ac+bc+c^2+cd)+(ad+bd+cd+d^2)$$
$$\left(\frac{-B}{A}\right)^2=2(ab+ac+ad)+2(bc+bd)+2(cd)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2=2(ab+ac+ad+bc+bd+cd)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2=2\left(\frac{C}{A}\right)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2-2\left(\frac{C}{A}\right)=a^2+b^2+c^2+d^2$$
$$\frac{B^2}{A^2}-\frac{2AC}{A^2}=a^2+b^2+c^2+d^2$$
$$a^2+b^2+c^2+d^2=\frac{B^2-2AC}{A^2}$$
My Question:

*

*Is my work correct ?

*Is it possible trying from $abcd=\frac{E}{A}$ ?

Thank you for your help and your time. God bless you.

Indeed, we know that
$$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd$$ and you find these sums from Vieta's formulas.

Solving from $abcd$ alone is obviously impossible, because there are infinitely many ways to obtain the same value from different $a,b,c,d$, and these result in different sums $a^2+b^2+c^2+d^2$. To make $a,b,c,d$ uniquely determined, you need more information, i.e. more polynomial coefficients.
(The complete discussion of why all other possible combinations of the polynomial coefficients do not work is out of my reach/time.) Note that you can't retrieve the four roots from less than four coefficients, but we are not looking for the individual roots but for the sum of squares.
