Engr2012 wrote:
Bunuel,
If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".
You are NOT correct.
If you square \(x = \sqrt{2x - 1}\) you get \(x^2 = 2x - 1\).
Next, if you substitute x=1 into \(x = \sqrt{2x - 1}\) the equation will hold true.
I think the confusion is between 2^0.5*x-1 versus (2x-1)^(0.5), in the 2nd case the problem and solution as stated is correct...