Question 48164This question is from textbook College Algebra
: Find all solutions of the system of equations
xy=24
2x^2-y^2+4 = 0
Thank you so very much!
This question is from textbook College Algebra
Answer by deepblue(1)   (Show Source):
You can put this solution on YOUR website! 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
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