+1348 The graphs of $x=y^4$ and $x+y^2=1-x$ intersect at two points. The distance between these points is of the form $u+v\sqrt2$, where $u$ and $v$ are integers. Find the ordered pair $(u,v)$.
+129895 Sub the first equation into the second and we have
y^4 + y^2 = 1 - y^4
2y^4 + y^2 - 1 = 0 factor this
(2y^2 -1) (y^2 + 1) = 0
The second factor has no real solution......set the first factor = 0
So
2y^2 - 1 = 0
2y^2 = 1
y^2 = 1/2
y = 1/sqrt (2) or y = -1/sqrt (2)
And x = y^4 = (1 / sqrt (2))^4 = 1/4
The distance between these points is just 1/sqrt (2) - -1/sqrt (2) = 2/sqrt (2) = sqrt (2) = 0 + 1sqrt (2)
(u,v) = (0, 1)
