Question

How do extraneous solutions arise from radical equations?

Answer 1

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.