WMC 2018 Pursuit - Pack 1, B2

Algebra Level 3

How many real solutions are there to the following system:

{ x 3 2 x y 2 = 81 2 x 2 y y 3 = 0 \begin{cases} x^3-2xy^2=81 \\ 2x^2y-y^3=0 \end{cases}

WMC Pursuit Questions


The answer is 3.

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1 solution

I got 3 \boxed 3 solution pairs ( x , y ) (x,y) . From

2 x 2 y y 3 = 0 y ( 2 x + y ) ( 2 x y ) = 0 \begin{aligned} 2x^2y - y^3 & = 0 \\ y(\sqrt 2x+y)(\sqrt 2 x - y) & = 0 \end{aligned}

{ y = 0 x = 3 3 3 y = 3 2 x = 3 y = 3 2 x = 3 \implies \begin{cases} y = 0 & \implies x = 3\sqrt[3]3 \\ y = 3\sqrt 2 & \implies x = - 3 \\ y = -3\sqrt 2 & \implies x = - 3 \end{cases}

@Chew-Seong Cheong , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 2 years, 8 months ago

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