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# help plz

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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).

Feb 15, 2019

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x = y^4      (1)      and     x + y^2 = 1      (2)

Sub   (1)  into (2)   and we have that

y^4 + y^2 = 1

Let  y^2  =  a      and we have

a^2 + a   =   1        complete the square on a

a^2 + a + 1/4   =  1 + 1/4       factor the left....simplify the right

(a + 1/2)^2  =  5/4           take both roots

a + 1/2   =   ±√5  /  2       subtract   1/2 from both sides

a =   ( - 1 ±√5 ) / 2

Since  y^2  must be positive, then  a  must be   ( √ 5 - 1 ) / 2    =  y^2      (3)

Using   x + y^2 = 1

x =  1 - y^2

x = 1 -  [  ( √ 5 - 1 ) / 2   ]  =    ( 3 - √ 5 ) / 2

These curves intersect at the same x values

So...the distance between the y values is the distance between the intersection points

So.....using (3)

y =  √  [ ( √ 5 - 1 ) / 2 ]       or y =  - √ [ ( √ 5 - 1 ) / 2 ]

So... the distance between these points is just

√  [ ( √ 5 - 1 ) / 2 ]   -  [  - √ [ ( √ 5 - 1 ) / 2 ]  ]  =

2 √  [ ( √ 5 - 1 ) / 2 ]   =

√ [ 4 ( √5 - 1 ) / 2 ]  =

√ [ 2√ 5 - 2 ]  =

√ [ - 2 + 2√ 5 ]   ⇒  (u, v)   =  (-2, 2)   Feb 15, 2019