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Question 65
If $$x^2-xy+y^2=2$$ and $$x^4+x^2y^2+y^4=6$$, then the value of $$x^2 + xy + y^2$$ is:
Solution
Given, $$x^2-xy+y^2 = 2$$
$$x^2+y^2 = 2+xy$$ --> (1)
Squaring on both sides
$$x^4+y^4+2x^2y^2 = 4+x^2y^2+4xy$$
$$x^4+x^2y^2+y^4 = 4+4xy$$
Given, $$x^4+x^2y^2+y^4 = 6$$
=> $$6 = 4+4xy$$
4xy = 2
xy = 0.5
From (1) , $$x^2+xy+y^2 = 2+xy+xy = 2+0.5+0.5 = 3$$
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