Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# 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.