If #x=2+sqrt(3)# and #y=2-sqrt(3)# then what is #x^2+y^2+xy# ?

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Answer 1 · 2015-09-04T18:49:48

15

#x^2+y^2+xy= (x+y)^2-xy#

#= (2+sqrt(3)+2-sqrt3)^2-[(2+sqrt3)(2-sqrt3)]#

#=16-(4-3)=15#

used: #(a+b)^2=a^2+b^2+2ab#
#(a+b)(a-b)=a^2-b^2#

Answer 2 · 2015-09-04T18:50:29

#15#

If #x=2+sqrt(3)#
#color(white)("XXXXXXXX")x^2=4+4sqrt(3)+3=color(red)( 7+4sqrt(3))#

If #y=2-sqrt(3)#
#color(white)("XXXXXXXX")y^2=4-4sqrt(3)+3= color(red)(7 - 4sqrt(3))#

and
#color(white)("XXXXXXXX")xy = 4-3 color(white)("XXXXX")=color(red)(1)#

Sum#=color(white)("XXXXXXXXXXXXXXXX")= color(blue)(15)#

Answer 3 · 2015-09-04T18:52:24

#x^2+y^2+xy = 15#

Use the difference of squares identity:

#a^2 - b^2 = (a+b)(a-b)# to calculate #xy#:

#xy = (2+sqrt(3))(2-sqrt(3)) = 2^2-sqrt(3)^2 = 4-3 = 1#

#color(white)()#
Also #(x+y)^2 = x^2+2xy+y^2#

So:

#x^2+y^2+xy = x^2+2xy+y^2 - xy#

#=(x+y)^2 - xy = 4^2-1 = 16-1 = 15#