Question

Is there anyone out there who knows the formula for the sum of the squares of the roots for quadratic, cubic, quartic, etc functions? I know the formulas for the more common ones like sum, product, sum taken a certain number at a time, sum of reciprocals, etc, but I don't know this one.

Answer 1

Also, I'm aware that all of them follow the law of signs, so if I can just get the quadratic one, I can adapt it to my own needs later.

Answer 2

I can tell you what it is for a quadratic, but it might be a little while longer before I can figure out the cubic and quartic versions. (Although it might be the obvious generalization of the quadratic)

Given a quadratic \[x^2+bx+c\]with roots \(r_1\) and \(r_2\), \[r_1^2+r_2^2=b^2-2c\]

Answer 3

For a cubic\[x^3+bx^2+cx+d\]with roots \(r_1,r_2,r_3\) you have that \[r_1^2+r_2^2+r_3^2=b^2-2c\]So I believe that this will generalize to all n-degree polynomials.

Answer 4

How would that work? I understand why all of the other ones work, but this one seems a bit strange to me...

Answer 5

Why the sum is \(b^2-2c\) or why it generalizes?

Answer 6

Either or I suppose? I just want a general idea for how you got that. Like, for example,
\[\text{Sum of Reciprocals: }\frac{constant}{a}\space (\text{Following sign rule})\]When you have this:
\[\frac{1}{r_{1}} + \frac{1}{r_{2}} = \frac{r_1 + r_2}{r_1r_2}\]which is the sum of the roots of the product of the roots. Can you give me an explanation like this or of the sort?

Answer 7

I mean constant over linear coefficient

Answer 8

I'll prove it for the quadratic polynomial. From that you should be able to see the process to find it for the cubic and further.

Answer 9

You're given a polynomial \((x-r_1)(x-r_2)=0\). Expand this out, and you get \[x^2-(r_1+r_2)x+r_1r_2=x^2+bx+c=0\]Do a little substitution, and you find that \(b=-(r_1+r_2)\), and \(c=r_1r_2\). Note that \[b^2=(r_1^2+2r_1r_2+r_2^2)\]Hence, \[b^2-2r_1r_2=b^2-2c=r_1^2+r_2^2\]It is from very similar arguments one can derive the formula for an arbitrary degree polynomial.

Answer 10

Oh I see.I just went back to an old algebra book (I forgot I had it) I use and they had this formula for the sum of the squares for quadratics. Is it any different?
\[(-\frac{b}{a})^{2} - 2(\frac{c}{a})\]

Answer 11

That's the same thing except it's for a polynomial \[ax^2+bx+c=0\]You have to first divide by \(a\) to get a coefficient of 1, and then you can apply the formula I derived above.

Answer 12

Oh, nevermind. I see it. Thank you :) I asked it before, but no one else knew how to get it

Answer 13

You're welcome. If you want, I could probably show you a proof of the formula for a polynomial of degree \(n\).

Answer 14

No, it's ok. I want to try it myself now. If I get stuck, I'll come back, but for now, it's ok :)