We have , or . By Vieta's formulas, x and y are the roots of the quadratic equation .

Results 1 to 4 of 4

- October 17th 2012, 06:35 AM #1

- Joined
- Oct 2012
- From
- England
- Posts
- 2

- October 17th 2012, 06:56 AM #2

- Joined
- Oct 2009
- Posts
- 5,572
- Thanks
- 789

## Re: s = x^2 + y^2, t = 2xy

We have , or . By Vieta's formulas, x and y are the roots of the quadratic equation .

- October 17th 2012, 07:13 AM #3

- Joined
- Oct 2012
- From
- England
- Posts
- 2

## Re: s = x^2 + y^2, t = 2xy

- October 17th 2012, 07:27 AM #4

- Joined
- Oct 2009
- Posts
- 5,572
- Thanks
- 789

## Re: s = x^2 + y^2, t = 2xy

Vieta's formulas for quadratic equations are easy to derive. If an equation has roots x and y, then by the polynomial remainder theorem (the link has a short proof). Expanding the right-hand side and equating the coefficients of and z in both sides, we get b = -(x + y) and c = xy. So, x and y are the roots of . You can express the coefficients of this equation through s and t.

Once you have a quadratic equation with roots x and y, they can be expressed through s and t using the quadratic formula.