+86 The graphs of \(x=y^4\) and \(x+y^2=1\) intersect at two points. The distance between these points is of the form \(\sqrt{u+v\sqrt5}\), where \(u\) and \(v\) are integers. Find the ordered pair \((u,v)\).
+9665 \(\begin{cases} x = y^4\\ x + y^2 = 1 \end{cases}\)
Substituting,
\(y^4 + y^2 - 1= 0\\ y^2 = \dfrac{-1 \pm \sqrt 5}2\text{(reject negative root)}\\ y^2 = \dfrac{\sqrt 5 - 1}2\\ y = \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\)
Since \(x + y^2 = 1\),
\(x = 1 - \dfrac{\sqrt 5 - 1}2 = \dfrac{3 -\sqrt 5}2\)
So the intersection points are \(\left(\dfrac{3 - \sqrt 5}2, \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\right)\).
The distance is \(2 \sqrt {\dfrac{\sqrt 5 - 1}2} = \sqrt{2(\sqrt 5 - 1)} = \sqrt{-2 + 2 \sqrt 5}\).
Then \((u,v) = (-2, 2)\).
