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Question Number 89187 by TawaTawa1 last updated on 16/Apr/20

Commented by jagoll last updated on 16/Apr/20

(√(8xy−1)) = t 8xy = t^2 +1 ⇒(√(8xy)) = (√(t^2 +1)) 4y = ((t^2 +1)/(2x)) (2x−1)^2 +4(√(t^2 +1)) = 4 ((t^2 +1)/(2x))− t = 1 ⇒ t^2 +1 = 2x(t+1)

8xy−1=t8xy=t2+1⇒8xy=t2+14y=t2+12x(2x−1)2+4t2+1=4t2+12x−t=1⇒t2+1=2x(t+1)

Commented by jagoll last updated on 16/Apr/20

4(√(t^2 +1)) = 4−(2x−1)^2 16(t^2 +1) = {4−(2x−1)^2 }^2

4t2+1=4−(2x−1)216(t2+1)={4−(2x−1)2}2

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