Asked by Bob
Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 62$.
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Answered by
GPT-5 mini
AI
From x+y=10 we have (x+y)^2 = x^2 + 2xy + y^2, so
100 = 62 + 2xy ⇒ 2xy = 38 ⇒ xy = 19.
Thus x and y are roots of t^2 - (x+y)t + xy = t^2 - 10t + 19 = 0.
Solve: t = (10 ± √(100 - 76))/2 = (10 ± √24)/2 = 5 ± √6.
So the ordered pairs are (5+√6, 5-√6) and (5-√6, 5+√6).
100 = 62 + 2xy ⇒ 2xy = 38 ⇒ xy = 19.
Thus x and y are roots of t^2 - (x+y)t + xy = t^2 - 10t + 19 = 0.
Solve: t = (10 ± √(100 - 76))/2 = (10 ± √24)/2 = 5 ± √6.
So the ordered pairs are (5+√6, 5-√6) and (5-√6, 5+√6).
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