How do extraneous solutions arise from radical equations?

In general, extraneous solutions arise when we perform non-invertible operations on both sides of an equation. (That is, they sometimes arise, but not always.)

Squaring (or raising to any other even power) is a non-invertible operation. Solving equations involving square roots involves squaring both sides of an equation.

Example 1 : To show the idea:
The equations: #x-1=4# and #x=5#, have exactly the same set of solutions. Namely: #{5}#.

Square both sides of #x=5# to get the new equation: #x^2=25#. The solution set of this new equation is; #{-5, 5}#. The #-5# is an extraneous solution introduced by squaring the two expressions

Square both sides of #x-1=4# to get #x^2-2x+1=16#
which is equivalent to #x^2-2x-15=0#.
and, rewriting the left, #(x+3)(x-5)=0#.
So the solution set is #{-3, 5}#.
This time, it is #-3#, that is the extra solution.

Example2 : Extraneous solution.
Solve #x=2+sqrt(x+18)#
Subtracting #2# from both sides: #x-2=sqrt(x+18)#

Squaring (!) gives #x^2-4x+4=x+18#
This requires, #x^2-5x-14=0#.
Factoring to get #(x-7)(x+2)=0#

finds the solution set to be #{7, -2}#.
Checking these reveals that #-2# is not a solution to the original equation. (It is a solution to the 3rd equation -- the squared equation.)

Example 3 : No extraneous solution.
Solve #sqrt(x^2+9)=x+3#
Squaring (!) gives, #x^2+9=(x+3)^2=x^2+6x+9#
Which leads to #0=6x# which has only one solution, #0# which works in the original equation.