If x+y=12 and xy =14,find the value of (x^2+y^2)
MANKOTIA:
x^2+y^2=116
Answers
Answered by
7
USING IDENTITY (a+b)^2=a^2+2ab+b^2
(x+y)^2=x^2+2xy+y^2
(12)^2=x^2+y^2+(2×14)
144=x^2+y^2+28
x^2+y^2=144-28=116
HOPE THIS WILL HELP U
(x+y)^2=x^2+2xy+y^2
(12)^2=x^2+y^2+(2×14)
144=x^2+y^2+28
x^2+y^2=144-28=116
HOPE THIS WILL HELP U
Answered by
2
(x+y)^2=x^2+y^2+2ab
(12)^2=x^2+y^2+2(14)
144=x^2+y^2+28
x^2+y^2=144-28
x^2+y^2=116
(12)^2=x^2+y^2+2(14)
144=x^2+y^2+28
x^2+y^2=144-28
x^2+y^2=116
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