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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 $(u,v)$.

 Aug 14, 2023
 #1
avatar+126978 
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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)

 

 

cool cool cool

 Aug 15, 2023

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