Asked by Scraffii
If (x+y)^2=168 and xy =4, then x^2 + y^2=?
Answers
Answered by
Henry
(x+y)^2 = 168
x^2 + 2xy + y^2 = 168
Replace xy with 4:
x^2 + 2*4 + y^2 = 168
x^2 + y^2 = 168 - 8 = 160.
x^2 + 2xy + y^2 = 168
Replace xy with 4:
x^2 + 2*4 + y^2 = 168
x^2 + y^2 = 168 - 8 = 160.
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