h ch Solutiony1 x2 12 y2 2x21 y1 exceeds y2 by 3 therefor

Solution

y1 = (x^2 - 1)^2

y2 = 2(x^2-1)

y1 exceeds y2 by 3

therefore ,

y1 = y2+3

(x^2 - 1)^2 = 2(x^2-1) + 3

x^4 + 1 -2x^2 = 2x^2 - 2 + 3

x^4 + 1 - 2x^2 = 2x^2 + 1

x^4 - 4x^2 = 0

taking x^2 as gcf

x^2( x^2 - 4) = 0

x^2 = 0

(x^2 - 4) = 0

x^2 = 4

x = +2 , -2

therefore, x = 0 , +2 , - 2