SOLUTION: Solve the quadratic system. Solve algebraically: 4x^2-9y^2=108 and xy=-12

Question 407387: Solve the quadratic system.
Solve algebraically: 4x^2-9y^2=108 and xy=-12
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
xy=-12

y=-12/x

4x^2-9y^2=108

4x^2-9(-12/x)^2=108

4x^2-9(144/x^2)=108

4x^4-1296=108x^2

4x^4-108x^2-1296=0

4z^2-108z-1296=0

Notice we have a quadratic equation in the form of where , , and

Let's use the quadratic formula to solve for z

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Rewrite as

Add to to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So the answers in terms of z are or

Now above, I let z = x^2. So x^2 = 36 and x^2 = -9

Solve for x in each equation to get: x = 6, x = -6, x = 3i, or x = -3i

So the four solutions for x are x = 6, x = -6, x = 3i, or x = -3i

Now plug all of these x values into y = -12/x to find the corresponding values of y. I'll let you do this.

In the end, you should have 4 ordered pairs.

If you need more help, email me at jim_thompson5910@hotmail.com

Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you

Jim

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