Question

Sum of the square of the roots of the cubic polynomials

Answer 1

Well, I recently proved a formula (at least, I think) to the sum of the inverse of the roots x1,x2,x3,…,xn∈C, and ≠0. It starts:

Let a polynomial P(x)=anxn+an−1xn−1+⋅⋅⋅+a1x+a0 of roots x1,x2,x3,…,xn∈C∗ and an≠0. So,

1x1+1x2+1x3+⋯+1xn=−a1a0,a0≠0.

Now, I am trying to prove another formula, the sum of the square of the roots, but I think it's getting pretty difficult to me.

Let x21+x22+x23+⋯+x2n=u. So, if x1⋅x2⋅x3⋯xn=(−1)na0an, then

x1=(−1)na0(x2⋅x3⋯xn)an,

and

x21=(−1)nx1a0(x2⋅x3⋯xn)an.

So,

x21+x22+⋯+x2n=u=(−1)nx1a0(x2⋅x3⋯xn)an+(−1)nx2a0(x1⋅x3⋯xn)an+⋯+(−1)nxna0(x1⋅x2⋯xn−1)an.

It can be written as

(−1)n⋅a0an(x1x2⋅x3⋯xn+x2x1⋅x3⋯xn+⋯+xnx1⋅x2⋯xn−1)=u

Answer 2

Step-by-step explanation:

Find the sum of the squares of the roots of the cubic equation x 3 + 3 x 2 + 3 x = 3 x^3 + 3x^2 + 3x = 3 x3+3x2+3x=3.

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Relation between coefficients and roots:

Root expression Equals to

p + q + r p+q+r p+q+r − b a -\frac{b}{a} −ab

p q + q r + r p pq+qr+rp pq+qr+rp

Hope its help...