SOLUTION: Solve the following system of nonlinear equations. 2xy=3 4x^2-8y^2=1

Question 169249: Solve the following system of nonlinear equations.
2xy=3
4x^2-8y^2=1
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Solve the following system of nonlinear equations.
2xy=3
4x^2-8y^2=1
--------------
From the 1st eqn, x = 3/2y
Sub for x into the 2nd eqn:
4(3/2y)^2 - 8y^2 = 1
9/(y^2) - 8y^2 = 1
Multiply by y^2
9 - 8y^4 = y^2
8y^4 + y^2 - 9 = 0
This is a quadratic in y^2

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=289 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1, -1.125. Here's your graph:

Factor:
(y^2 - 1)*(8y^2 + 9) = 0
y^2 = 1, -9/8
y = +1, -1
y = +i*(3/4)sqrt(2)
y = -i*(3/4)sqrt(2) (i = sqrt(-1) )

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