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

1 Answer

Jul 19, 2016

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

Explanation:

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]))}#

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