Question 765632
x(square) + y(square) = 4
xy = 1 
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Solve for "y":
y = 1/x
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Substitute for "y" and solve for "x":
x^2 + (1/x)^2 = 4
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x^4 + 1 = 4x^2
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x^4 - 4x^2 + 1 = 0
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x^2 = [4 +- sqrt(16-4)]/2
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x^2 = [4+- 2sqrt(3)]/2
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x^2 = 2 +- sqrt(3)
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x = sqrt[2+sqrt(3)] or x = -sqrt[2+sqrt(3)]
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Solve for "y"
y = 1/sqrt[2+sqrt(3)] or y = -1/sqrt[2+sqrt(3)]
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Cheers,
Stan H.
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