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Mathematics 13 Online
OpenStudy (anonymous):

If (x+y)^2=121 and xy=28 find the value of x^2+y^2

OpenStudy (anonymous):

Based on property: (a+b)^2 = a^2 + 2ab + b^2 thus (x+y)^2 = 121 = x^2 + 2xy + y^2 121 - 2xy = x^2 + y^2 121 - 2(28) = x^2 + y^2 = ?

OpenStudy (anonymous):

65! :D thanks!!! :)

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