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

• Oct 17th 2012, 05:35 AM
patpatpat
s = x^2 + y^2, t = 2xy
I need to find x and y in terms of t and s.

Help?
• Oct 17th 2012, 05:56 AM
emakarov
Re: s = x^2 + y^2, t = 2xy
We have $\displaystyle (x + y)^2 = s + t$, or $\displaystyle x + y = \sqrt{s + t}$. By Vieta's formulas, x and y are the roots of the quadratic equation $\displaystyle z^2-z\sqrt{s+t}+t/2=0$.
• Oct 17th 2012, 06:13 AM
patpatpat
Re: s = x^2 + y^2, t = 2xy
Quote:

Originally Posted by emakarov
We have $\displaystyle (x + y)^2 = s + t$, or $\displaystyle x + y = \sqrt{s + t}$. By Vieta's formulas, x and y are the roots of the quadratic equation $\displaystyle z^2-z\sqrt{s+t}+t/2=0$.

Excuse the ignorance, but I'm unfamiliar with Vieta's Formulas. I understand that $\displaystyle (x + y)^2 = s + t$ - didn't realise this forum had tex in bb, that's pretty cool - but how do I go about the next step? I'm still at a loss at how to attain my final two x and y equations that I need.
• Oct 17th 2012, 06:27 AM
emakarov
Re: s = x^2 + y^2, t = 2xy
Quote:

Originally Posted by patpatpat
Excuse the ignorance, but I'm unfamiliar with Vieta's Formulas.

Vieta's formulas for quadratic equations are easy to derive. If an equation $\displaystyle z^2+bz+c=0$ has roots x and y, then $\displaystyle z^2+bz+c=(z-x)(z-y)$ by the polynomial remainder theorem (the link has a short proof). Expanding the right-hand side and equating the coefficients of $\displaystyle z^2$ and z in both sides, we get b = -(x + y) and c = xy. So, x and y are the roots of $\displaystyle z^2 -(x+y)z+(xy)z=0$. You can express the coefficients of this equation through s and t.

Quote:

Originally Posted by patpatpat
how do I go about the next step? I'm still at a loss at how to attain my final two x and y equations that I need.

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