Let u=x-1. We can then rewrite the left hand side of the equation as
sqrt(u+4-4sqrt(u)) + sqrt(u+9-6sqrt(u))
=sqrt((sqrt(u)-2)^2) + sqrt((sqrt(u)-3)^2)
=|sqrt(u)-2| + |sqrt(u)-3|
Note the presence of sqrt(u) in the equation and that we are only looking for real values, so we have the restriction u >= 0. With that, we will now consider all remaining cases:
Case 1: 0 <= u <= 4
|sqrt(u)-2| + |sqrt(u)-3| = 1
=> 2-sqrt(u) + 3-sqrt(2) = 1
=> -2sqrt(u) = -4
=> sqrt(u) = 2
=> u = 4
Thus u=4 is the only solution in the interval [0, 4]
Case 2: 4 <= u <= 9
|sqrt(u)-2| + |sqrt(u)-3| = 1
=> sqrt(u)-2 + 3 - sqrt(u) = 1
=> 1=1
As this is a tautology, every value in [4, 9] is a solution.
Case 3: u >= 9
|sqrt(u)-2| + |sqrt(u)-3| = 1
=> sqrt(u) - 2 + sqrt(u) - 3 = 1
=> 2sqrt(u) = 6
=> sqrt(u) = 3
=> u = 9
Thus u = 9 is the only solution in the interval [9, oo)
Taken together, we have [4, 9] as the solution set for real values of u. Substituting in x = u+1, we arrive at the final solution set x in [5, 10]
Looking at the graph of the left hand side, this matches with what we would expect:
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