Substitution instead of Squaring
Let a = sqrt(x-1) and b = sqrt(x+1)
Domains: x >= 1, a >= 0, b >= √2
x = a^2 + 1 = b^2 - 1
x = (a^2 + b^2)/2
a^2 + 1 + a + b + ab - 4 = 0
(a^2 + b^2)/2 + a + b + ab - 4 = 0
(a^2 + 2ab + b^2)/2 + a + b - 4 = 0
(a+b)^2 + 2(a+b) - 8 = 0
(a+b)^2 + 2(a+b) + 1 = 9
(a+b+1)^2 = 9
a+b+1 = ±3
a + b = {-4,2}
reject -4 because from the domains: a + b >= √2
So: a + b = 2
Recall x = a^2 + 1 = b^2 - 1
a^2 - b^2 = -2
(a+b)(a-b) = -2
(a+b) = 2
(a-b) = -1
(a,b) = (1/2, 3/2)
sqrt(x-1) = 1/2
sqrt(x+1) = 3/2
x-1 = 1/4
x+1 = 9/4
x = 5/4
|
|
Posted by Larry
on 2023-12-04 09:19:55 |
Solution
