Friday, October 28, 2011

What is the real solution of the system x^2+y^2=16, xy=3 ?

We have the system of equations x^2 + y^2 = 16 and xy = 13
to solve for x.


xy = 3 => x =
3/y...(1)


Substitute this in x^2 + y^2 =
16


=> (3/y)^2 + y^2 =
16


=> 9 + y^4 =
16y^2


=> y^4 - 16y^2 + 9 =
0


y^2 = 16/2 + [16^2 -
36]/2


=> 8 + (sqrt
220)/2


=> 8 + sqrt
55


and y^2 = 8 - sqrt 55


y =
sqrt (8 + sqrt 55) and y = sqrt(8 - sqrt 55)


x = 3/(sqrt (8
+ sqrt 55)) and x = 3/(sqrt(8 - sqrt 55))


Also in (1) we
could have substituted x for y. This gives us four solutions for x and
y:


(sqrt (8 + sqrt 55), 3/(sqrt (8 + sqrt 55))), (sqrt (8 -
sqrt 55), 3/(sqrt (8 - sqrt 55))), (3/(sqrt (8 + sqrt 55)), sqrt (8 + sqrt 55)) and
(3/(sqrt (8 - sqrt 55)), sqrt (8 - sqrt
55))


The solutions of the equations are (sqrt
(8 + sqrt 55), 3/(sqrt (8 + sqrt 55))), (sqrt (8 - sqrt 55), 3/(sqrt (8 - sqrt 55))),
(3/(sqrt (8 + sqrt 55)), sqrt (8 + sqrt 55)) and (3/(sqrt (8 - sqrt 55)), sqrt (8 - sqrt
55))

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