How do extraneous solutions arise from radical equations?
In general, 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.