If (x-y)^2-(x+y)^2=-20, determine the value of xy

asked Jun 24, 2016 80.4k views

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$(x-y)^2-(x+y)^2=-20\\ x^2-2xy+y^2-(x^2+2xy+y^2)=-20\\ x^2-2xy+y^2-x^2-2xy-y^2=-20\\ -4xy=-20\\ xy=5$

Ji answered Jun 25, 2016

by Ji

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(x-y)^2-(x+y)^2= -20
x^2-2xy+y^2-(x^2 + 2xy + y^2)= -20
x^2 -2xy +y^2 -x^2 -2xy -y^2= -20
-4xy= -20
xy = -20/ -4
xy = 5

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