Finding the sum of squares of roots of a quartic polynomial.

Solution 1:

We have that

$$(x-a)(x-b)(x-c)(x-d)=$$ $$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$

then by

• $S_1=a+b+c+d$
• $S_2=ab+ac+ad+bc+bd+cd$
• $S_3=abc+abd+acd+bcd$
• $S_4=abcd$

$$a^2+b^2+c^2+d^2=S_1^2-2S_2$$

and more in general by Newton's sums we have that

• $P_1=a+b+c+d=S_1$
• $P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2$
• $P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3$
• $P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4$

Solution 2:

Hint:

\begin{align} \sum_{i=1}^4a_i^2 &= \left(\sum_{i=1}^4a_i \right)^2-2\sum_{i< j}a_ia_j \end{align}

Also Vieta's formula might help.

Solution 3:

Hint:

$$r_0^2+r_1^2+r_2^2+r_3^2=(r_0+r_1+r_2+r_3)^2-2(r_0r_1+r_0r_2+r_0r_3+r_1r_2+r_1r_3+r_2r_3)$$

and Vieta.