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find two numbers whose sum is 13 and the sum of whose square is 145
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x+y=13 x^2+y^2=145
x = 13-y (13-y)^2+y^2=145
solve the 2nd equation: 169-26y+y^2+y^2=145 2y^2 -26y + 169 - 145 = 0
2y^2 - 26y + 24 = 0 y^2 - 13y + 12 = 0
D = 169 - 48 = 121 y1 = (13 - 11)/2 = 1 y2 = (13+11)/2 = 12 So now you have two y values..
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Find the two x values: x1 = 13-y1 = 13-1 = 12 x2 = 13-y2 = 13 -12 = 1
So apparently the two numbers u need are 12 and 1..
very good bahrom
did you have the same answer??...myininaya??
yes his answers are glamorous
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ok..
Heres another solution that uses the symmetry of the problem:
very nice joe
tnx!!
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