Answer to Question #125928 in Algebra for Ojugbele Daniel

Question #125928
x⁴+x²y²+y⁴ = 133
x²+xy+y² = 19

Solve for x and y
1
Expert's answer
2020-07-12T17:37:53-0400

Let x2+y2=u,xy=v.x^2+y^2=u, xy=v. Then x4+x2y2+y4=(x2+y2)2x2y2=u2v2x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=u^2-v^2


u2v2=133u^2-v^2=133u+v=19u+v=19

(uv)(u+v)=133(u-v)(u+v)=133u+v=19u+v=19

uv=7u-v=7u+v=19u+v=19

u=13u=13v=6v=6

Hence


x2+y2=13x^2+y^2=13xy=6xy=6

(x+y)22xy=13(x+y)^2 -2xy=13xy=6xy=6

(x+y)2=25(x+y)^2=25xy=6xy=6

If x+y=5x+y=5


(2,3),(3,2)(2, 3), (3, 2)

If x+y=5x+y=-5


(3,2),(2,3)(-3, -2), (-2, -3)

Therefore


(3,2),(2,3),(2,3),(3,2)(-3, -2), (-2, -3), (2, 3), (3, 2)


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