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# Question

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Find the product of the y-coordinates of all the distinct solutions (x,y) for the two equations \(y=x^2-8\) and \(y^2=-5x+44.\)

Aug 25, 2018

#1
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y = x^2  - 8   ⇒  y^2  = x^4 - 16x^2 + 64   (1)

y^2  = -5x + 44    (2)

Set (1)  = (2)   and solve for x

x^4 - 16x^2 + 64  = -5x + 44

x^4 - 16x^2  + 5x  + 20  = 0

x^2 (x^2 - 16) + 5( x +  4)  = 0

x^2 (x + 4)(x - 4) + 5 (x + 4)  = 0

(x + 4) [ x^2 ( x - 4) + 5 ]   =  0

( x + 4) [ x^3 - 4x^2 + 5 ] = 0

x = -4  is one solution

Also  x = -1  is another solution

To find the remaining polynomial  use synthetic division

-1 [ 1  - 4    0     5]

-1    5    -5

____________

1   -5    5     0

So...the remaining polynomial is  x^2 - 5x + 5....set to  0

x^2-5x + 5  = 0

x^2 - 5x + 25/4  = -5 + 25/4

(x - 5/2)^2  = 5/4     take both roots

x - 5/2  = ± √5 /2

x = 5/2 + √5/2    or     5/2 - √5/2

When  x  = -4, y= (-4)^2 - 8   = 8

x = -1 , y = (-1)^2 - 8  = -7

x = 5/2 + √5/2,   y = (5//2  + √5/2)^2  - 8  =  (1/2)(5√5 - 1)= (1/2)(√125 - 1)

x = 5/2  - √5/2, y = (5/2 - √5/2)^2  - 8  = (-1/2)(5√5 + 1)  = (-1/2)(√125 + 1)

So....the product of the y coordinates is

( 8 * -7 * (1/2)*(-1/2)(√125 - 1)(√125 + 1 ] =

[ -56 * (-1/4) * (125 - 1) ] =

[ 14 * 124 ]  =

1736   Aug 25, 2018
#2
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Thank you, CPhill!

mathtoo  Aug 25, 2018
#3
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No prob, Mathtoo....!!!   CPhill  Aug 25, 2018