#x=1+sqrtx# if and only if
#x-1=sqrtx# which is true only if
#(x-1)^2=(sqrtx)^2#
So we need to solve #(x-1)^2=(sqrtx)^2# and check for extraneous solutions.
#(x-1)^2=(sqrtx)^2#
#x^2-2x+1 = x#
#x^2 - 3x +1=0#
By completing the square of the quadratic formula, we get:
#x_1=(3 + sqrt5)/2#, and #x_2=(3 - sqrt5)/2#
Checking #x_1#:
Note that #2 < sqrt 5 < 3#, so the solution #(3+sqrt5)/2# is greater than #1#, so
#x_1 -1 = (3+sqrt5)/2 -1# is positive and a permitted value for the principal square root: #sqrt(x_1)#
Checking #x_2#:
Note that #2 < sqrt 5 < 3#,
so #-3 < - sqrt5 < -2#. (multiply each part by -1, reversing the inequalities)
Therefore, #0 < 3-sqrt5 < 1# (Add 3 to each part of the inequality)
the solution #x_2# is #0 < (3-sqrt5)/2 < 1/2#
Now, when we try to make #x_2 -1# equal to a principal square root (#sqrt(x_2)#), we cannot, because #x_2 - 1# is negative.
#x_2# is an extraneous solution. It solves #x^2-2x+1 = x# , but not #x-1=sqrtx#., so it also does not solve #x=1+sqrtx#.
