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

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Question 407387: Solve the quadratic system.
Solve algebraically: 4x^2-9y^2=108 and xy=-12
Answer by jim_thompson5910(35256) About Me  (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 az%5E2%2Bbz%2Bc where a=4, b=-108, and c=-1296


Let's use the quadratic formula to solve for z


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%28-108%29+%2B-+sqrt%28+%28-108%29%5E2-4%284%29%28-1296%29+%29%29%2F%282%284%29%29 Plug in a=4, b=-108, and c=-1296


z+=+%28108+%2B-+sqrt%28+%28-108%29%5E2-4%284%29%28-1296%29+%29%29%2F%282%284%29%29 Negate -108 to get 108.


z+=+%28108+%2B-+sqrt%28+11664-4%284%29%28-1296%29+%29%29%2F%282%284%29%29 Square -108 to get 11664.


z+=+%28108+%2B-+sqrt%28+11664--20736+%29%29%2F%282%284%29%29 Multiply 4%284%29%28-1296%29 to get -20736


z+=+%28108+%2B-+sqrt%28+11664%2B20736+%29%29%2F%282%284%29%29 Rewrite sqrt%2811664--20736%29 as sqrt%2811664%2B20736%29


z+=+%28108+%2B-+sqrt%28+32400+%29%29%2F%282%284%29%29 Add 11664 to 20736 to get 32400


z+=+%28108+%2B-+sqrt%28+32400+%29%29%2F%288%29 Multiply 2 and 4 to get 8.


z+=+%28108+%2B-+180%29%2F%288%29 Take the square root of 32400 to get 180.


z+=+%28108+%2B+180%29%2F%288%29 or z+=+%28108+-+180%29%2F%288%29 Break up the expression.


z+=+%28288%29%2F%288%29 or z+=++%28-72%29%2F%288%29 Combine like terms.


z+=+36 or z+=+-9 Simplify.


So the answers in terms of z are z+=+36 or z+=+-9


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