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

What is the sum of the squares of the roots of $x^4 - 8x^3 + 16x^2 - 11x + 5$ ?

This question is from the 2nd qualifying round of last year's Who Wants to be a Mathematician high school competition which can be seen here:

I know the answer (32) because that is also given in the link, and I have checked by brute force that the given answer is correct.

However, I have made no progress at all in figuring out how to calculate the sum of squares of the roots - either before or after knowing the answer! I was expecting there to be a nice "trick" analagous to the situation if they had given a quadratic and asked the same question -- in that case I know how to get the sum and product of the roots directly from the coefficients, and then a simple bit of algebraic manipulation to arrive at the sum of squares of the roots.

In this case (the quartic) I have no idea how to approach it, and I have not spotted any way to simplify the problem (e.g. I cannot see an obvious factorisation, which might have helped me).

I've looked on the web at various articles which dicuss the relationships between the coefficients of polynomials and their roots and - simply put - I found nothing which gave me inspiration for this puzzle.

Given the audience for this test, it should be susceptible to elementary methods ... I would appreciate any hints and/or solutions!

Thank you.

• $(\sum a_i)^2 = \sum a_i^2 + \sum_\limits{i\neq j} a_ia_j$ – AnotherJohnDoe Sep 17 '18 at 13:20
• This is a wonderful post, a rare essay for a new contributor. Since finding the solution is in this case more important than having it, here is a hint: Use the relations of Vieta for the four roots $a,b,c,d$ - say - of the given equation, which show how to extract the elementary symmetric polynomials, first two being $\sum a =a+b+c+d$ and $\sum ab=ab+ac+ad+bc+bd+cd$. Can you now find $(a+b+c+d)^2$? en.wikipedia.org/wiki/Vieta%27s_formulas – dan_fulea Sep 17 '18 at 13:21
• Thank you! This, and the answers posted soon afterwards were very helpful. – BBO555 Sep 17 '18 at 13:55

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

• $P_1=a+b+c+d=S_1$
• $P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2$
• $P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3$
• $P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4$
• Thanks for this very quick and helpful answer ! – BBO555 Sep 17 '18 at 13:56
• You are welcome! Bye – gimusi Sep 17 '18 at 14:21

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.

• Yes indeed - thank you! – BBO555 Sep 17 '18 at 13:57

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.

• Thank you! It appears the solution was much closer to hand than I realised at first ... – BBO555 Sep 17 '18 at 13:57
• @BBO555: maths are full of tricks... – Yves Daoust Sep 17 '18 at 15:39