Whole numbers are integers. a.always b.sometimes c.never

The answer to the question is always.

Part 1. create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution.
model a √ (x+b) + c = d
use a constant in place of each variable a, b, c, and d. you can use positive and negative constants in your equation.
a) a = -2, b = 3, c = -5, d = 7
=> -2√(x + 3) - 5 = 7
isolate the radical: - 2√(x+3) = 5 + 7
=> - 2√(x+3) = 12
=> √(x + 3) = -12/2
=> √x+3) = -6
square both sides
[√(x+3)]^2  =(-6)^2
=> x+3 = 36
=> = 36 - 3
=> x = 33
verify the validity of the solution, replacing 33 for x in the original equation:
=> -2√(33+ 3) - 5 = 7
=> -2 √(36) - 5 = 7
=> -2(6) - 5 = 7
=> -12 - 5 = 7
=> -17 = 7
which is not true. so, by solving the equation we obtained a solution which is no really a solution. that is what is called an extraneous solution, because it does not satisfy the equation, which means that you must discard that solution. in this case the conclusion is that the equation has not solution.
you could have prediceted it when you got √(x+3) = -6, given that the square root cannot lead to a negative number.
the reason to obtain the extraneous solution is that by squaring both sides you forced them to be positive.
that is why you must always verify the solutions of a radical equation and discard the extraneous ones.
b) a = 2, b = 3, c = -5, d = 7
=> 2√(x + 3) - 5 = 7
isolate the radical =>
2√(x + 3)  = 12
=> √(x+3) = 12 / 2
=> √(x + 3) = 6
square both sides =>
x + 3 = 36
=> x = 36 - 3
=> x = 33
now replace the value found to verify the solution:

=> 2√(x + 3) - 5 = 7
=> 2 √(33 +3) - 5 = 7
=> 2√(36)  - 5 = 7
=> 2(6) - 5 = 7
=> 12 - 5 = 7
=> 7 = 7
which proves that the solution is right.
part 2 and part three were answered in part one.
remember, some radical equations may have extraneous solutions because when you square the the sides you force to positive results which is not always valid.