Question 48164
HI freind,

these equations generally lead to quadratic equations

given xy=24 and 2x^2-y^2+4=0

in the second eqn substitute y=24/x
u get,

2x^2-(24/x)^2+4=0

simplifying we get 2x^4+4x^2-(24)^2=0

now assume x^2=p and substitute in teh above eqn we get

2p^2+4p-(24)^2=0

which is a normal quadratic equation

whose solutions are
p=16 and p=-18

case 1:x^2=p=16

=>  x=+sqrt(p) or - sqrt(p) =+4 or -4

case 2:when x^2=p=-18

x=- sqrt(-18)or + sqrt(-18)
   =-4.24i  or   4.24i  (we know sqrrt(-1)=i)

hence we have four slutions

two are real and otehr two are complex!

now having got x , y=24/x

so values of y follow (but take care! x=0 can never be a solution! if x=0 then y=24/x becomes 24/0....which is undefined!)

hope i solved ur question!

take care