+36 The distance between the two intersections of \(x=y^4\) and \(x+y^2=1\) is \(\sqrt{u+v\sqrt5}\). Find the ordered pair, \((u,v)\).
+6251 \(x = y^4 \text{ and } x = 1-y^2\\ y^4 = 1-y^2\\ y^4 +y^2 -1 = 0\\ y^2 = \dfrac{-1\pm \sqrt{1+4}}{2} = \dfrac{-1\pm \sqrt{5}}{2}\\ y = \pm \sqrt{\dfrac{-1\pm \sqrt{5}}{2}}\)
sorting through these possible solutions we find the only two real ones are
\(y = \pm \sqrt{\dfrac{-1+\sqrt{5}}{2}}\\ \text{both of these values of }y \text{ produce the same value of x}\\ \text{and thus the distance between them is }\\ \Delta y= 2 \sqrt{\dfrac{-1+\sqrt{5}}{2}} = \sqrt{-2 + 2\sqrt{5}}\\ (u,v) = (-2,2)\)
