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Math Help - s = x^2 + y^2, t = 2xy

  1. #1
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    s = x^2 + y^2, t = 2xy

    I need to find x and y in terms of t and s.

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

    We have (x + y)^2 = s + t, or x + y = \sqrt{s + t}. By Vieta's formulas, x and y are the roots of the quadratic equation z^2-z\sqrt{s+t}+t/2=0.
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  3. #3
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    Re: s = x^2 + y^2, t = 2xy

    Quote Originally Posted by emakarov View Post
    We have (x + y)^2 = s + t, or x + y = \sqrt{s + t}. By Vieta's formulas, x and y are the roots of the quadratic equation z^2-z\sqrt{s+t}+t/2=0.
    Excuse the ignorance, but I'm unfamiliar with Vieta's Formulas. I understand that (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.
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  4. #4
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    Re: s = x^2 + y^2, t = 2xy

    Quote Originally Posted by patpatpat View Post
    Excuse the ignorance, but I'm unfamiliar with Vieta's Formulas.
    Vieta's formulas for quadratic equations are easy to derive. If an equation z^2+bz+c=0 has roots x and y, then 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 z^2 and z in both sides, we get b = -(x + y) and c = xy. So, x and y are the roots of 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 View Post
    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.
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