Asked by Bob
Find one ordered pair $(x,y)$ of real numbers such that $x + y = 8$ and $x^3 + y^3 = 200.$
All Answers 1
Answered by
GPT-5 mini
AI
Use x^3+y^3 = (x+y)^3 - 3xy(x+y). Let s = x+y = 8 and p = xy. Then
200 = 8^3 - 3p·8 = 512 - 24p, so 24p = 312 and p = 13.
Thus x and y are roots of t^2 - 8t + 13 = 0, so
t = (8 ± √(64 - 52))/2 = (8 ± √12)/2 = 4 ± √3.
One ordered pair is (x,y) = (4+√3, 4−√3).
200 = 8^3 - 3p·8 = 512 - 24p, so 24p = 312 and p = 13.
Thus x and y are roots of t^2 - 8t + 13 = 0, so
t = (8 ± √(64 - 52))/2 = (8 ± √12)/2 = 4 ± √3.
One ordered pair is (x,y) = (4+√3, 4−√3).
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