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Halp

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There is a unique pair of positive real numbers satisfying the equations $x^4-6x^2y^2+y^4 = 8 \qquad\mbox{and}\qquad x^3y-xy^3 = 2\sqrt{3}.$Determine $x$, writing your answer in the form $a\cos\theta$, with $\theta$ in degrees.

I use WA and the x value is $\sqrt(2+\sqrt(3))$

how do I convert to a*cos(theta)

Jun 5, 2021
edited by Guest  Jun 5, 2021

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I graphed it and I see 4 real solutions.

Oh, sorry,   x and y must both be positive

https://www.desmos.com/calculator/a1oy7stjhb

Jun 5, 2021
edited by Melody  Jun 5, 2021