若(2x^2+2y^2)(x^2-2+y^2)-16=0,求x^2+y^2的值

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若(2x^2+2y^2)(x^2-2+y^2)-16=0,求x^2+y^2的值
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若(2x^2+2y^2)(x^2-2+y^2)-16=0,求x^2+y^2的值
若(2x^2+2y^2)(x^2-2+y^2)-16=0,求x^2+y^2的值

若(2x^2+2y^2)(x^2-2+y^2)-16=0,求x^2+y^2的值
设x^2+y^2=z
则原式＝2z(z-2)-16=0,
z^2-2z-8=0,
(z-1)^2=9,
z-1=3,z-1=-3;
z=4,z=-2,
又x^2+y^2>=0,则x^2+y^2=4

令t=x^2+y^2
则原式可化简为t（t-2）=8
t=4或-2（舍去
所以x^2+y^2=4

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