Find all real solutions of the system
x2+y2+[2x y/(x+y)]=1
| ___ Öx+y | =x2-y |
From the last part, we have
x+y = x4 - 2x2 y + y2
This we can rearrange into a quadratic in y, and we get
y2 + -(2x2 +1)y + (x4 -x) =
0.
Solving this by the normal methods, a miracle happens, and we see
that
2y = (2x2 +1) ± ( (1+2x2
)2 - 4(x4 -x) )1/2
2y = 2x2 +1 ± ( 1 + 4x2 +
4x4 - 4x4 + 4x )1/2
= 2x2 + 1 ± (2x+1)
So y is x2 + x + 1 or x2 -x.
Each of these may be quickly substituted into the first equation,
and the resulting equations can be factorised. One leads to the
quartic
-3 + 2x + 2x2 - 2x3 + x4
=0
of which ±1 are roots, and the others may be found by
dividing through by x2 -1.
The other case leads to the slightly more awkward equation
4x + 10x2 + 14x3 + 9x4 +
4x5 + x6 = 0
which is mildly awkward; 0 and -2 are obviously roots, and this
leaves us with a mildly nasty quartic, 2 + 4x + 5x2 +
2x3 + x4
However, as soon as we try to depress this by substituting x = z
- 1/2, to get rid of the cubic term, the whole thing falls to
pieces and we are left with a quadratic in z2 ,
z4 + 7/2 z2 + 17/16 = 0, which isn't
difficult to solve. (Of course this will have only complex roots
for z anyway, which is a bit irritating.)
This will give you 10 solutions for x and y. Six are complex and
can be ignored immediately, and the remainder are easily checked
to see if they fit the second equation (we might have introduced
spurious solutions in the squaring process).
In fact, we are left with only two solutions: (x,y) = (1,0) and
(-2,3).
David