Hello
Solve this equation and find the value of " x " and " y " respectively .
Thanks ×D
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x^4 +x^2y^2 +y +y^4 = (x^2 +y^2)^2 x^2y^2 =(x^2 +y^2 -xy)(x^2 +y^2 +xy)
x^2 -xy+y^2 =19
x^2+xy+ y^2 = 49
2x^2 +2y^2 =68
x^2 +y^2 =34
34 -xy =19
(x+y)^2 -2xy =34
xy= 15
(x+y)^2 -30=34
(x+y)^2 =64
xy = 15
x+y =8
x^2 -8x +15=0
x=3 x=5
y= 5 y=3
xy = 15
x+y = -8
x^2+8x+15=0
x= -5 x= -3
y = -3 y = -5
x^2 -xy+y^2 =19
x^2+xy+ y^2 = 49
2x^2 +2y^2 =68
x^2 +y^2 =34
34 -xy =19
(x+y)^2 -2xy =34
xy= 15
(x+y)^2 -30=34
(x+y)^2 =64
xy = 15
x+y =8
x^2 -8x +15=0
x=3 x=5
y= 5 y=3
xy = 15
x+y = -8
x^2+8x+15=0
x= -5 x= -3
y = -3 y = -5
ria113:
wow yaar !! super..
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