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Sunday, April 24, 2016

2016/034) Show that x^2-xy+y^2 - 4x - 4y + 16=0 has only one solution in real x,y. find it

 we have multiplying by 2
2x^2-2xy+2y^2 - 8x - 8y + 32=0
or (x^2 - 2xy + y^2) + (x^2-8x + 16) + (y^2 - 8y + 16) = 0
or (x-y)^2 +(x-4)^2 + (y-4)^2=0
hence we have each of the squares zero giving x=y=4 the only solution

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