SOLUTION: Find the real solutions, if any, of each system. {3m^2-2n^2= 9 {m^2-n^2= -1 The answers for this are (SquareRoot of 11, 2(Squareroot of 3)) (SquareRoot of 11, -2(Squarero

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the real solutions, if any, of each system. {3m^2-2n^2= 9 {m^2-n^2= -1 The answers for this are (SquareRoot of 11, 2(Squareroot of 3)) (SquareRoot of 11, -2(Squarero      Log On


   

Question 30021: Find the real solutions, if any, of each system.
{3m^2-2n^2= 9
{m^2-n^2= -1
The answers for this are
(SquareRoot of 11, 2(Squareroot of 3))
(SquareRoot of 11, -2(Squareroot of 3))
(-SquareRoot of 11, 2(Squareroot of 3))
(-SquareRoot of 11, -2(Squareroot of 3))
I just need to know how to go about solving this.
Thankyou
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+3m%5E2-2n%5E2=+9+ and +m%5E2-n%5E2=+-1+. Scale up the second one by multiplying it by 3.

+3m%5E2-2n%5E2=+9+ --eqn1
+3m%5E2-3n%5E2=+-3+ --eqn2

Now subtract it from the first equation, leaving:
+n%5E2+=+12+

so +n+=+sqrt%2812%29+ or +n+=+-+sqrt%2812%29+
--> +n+=+sqrt%284%2A3%29+ or +n+=+-+sqrt%284%2A3%29+
--> +n+=+sqrt%284%29sqrt%283%29+ or +n+=+-+sqrt%284%29sqrt%283%29+
--> +n+=+2sqrt%283%29+ or +n+=+-+2sqrt%283%29+

when +n+=+sqrt%2812%29+, we get +m%5E2+-+12+=+-1+ --> +m%5E2+=+11+
so, m+=+sqrt%2811%29+ or m+=+-+sqrt%2811%29+

Similar result obtained for +n+=+-+sqrt%2812%29+

jon.