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Find all pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).

 Feb 5, 2021
 #1
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From the given equations, x + y = 10 and xy = 14.  Then by Vieta's formulas, x and y are the roots of t^2 - 10t + 14 = 0.  These roots are 5 + sqrt(11) and 5 - sqrt(11), so the solutions are (5 + sqrt(11), 5 - sqrt(11)), (5 - sqrt(11), 5 + sqrt(11)).

 Feb 5, 2021
 #2
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I believe you could just substitute and solve. I am achieving (5 - sqrt3, 5 + sqrt3) and (5 + sqrt3, 5 - sqrt3).

 Feb 6, 2021

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