How do you solve $2x^2-3y^2+4=0$ and $xy=24$ ?

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Answer 1 · 2016-07-19T13:31:13

${x = pm sqrt[sqrt[865]-1], y =pm sqrt( 2/3 (1 + sqrt[865]))}$

Calling $x_2=x^2$ and $y_2=y^2$ we have
the equivalent system

${(2 x_2-3y_2+4=0), (x_2 y_2 = 24^2) :}$

Solving for $x_2,y_2$ we obtain the only feasible solution

${x2 = sqrt[865]-1, y2 = 2/3 (1 + sqrt[865])}$

then

${x = pm sqrt[sqrt[865]-1], y =pm sqrt( 2/3 (1 + sqrt[865]))}$

How do you solve 2x^2-3y^2+4=02x^2-3y^2+4=0 and xy=24xy=24?