# How do extraneous solutions arise from radical equations?

Mar 8, 2015

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: $\left\{5\right\}$.

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

Square both sides of $x - 1 = 4$ to get ${x}^{2} - 2 x + 1 = 16$
which is equivalent to ${x}^{2} - 2 x - 15 = 0$.
and, rewriting the left, $\left(x + 3\right) \left(x - 5\right) = 0$.
So the solution set is $\left\{- 3 , 5\right\}$.
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 ${x}^{2} - 4 x + 4 = x + 18$
This requires, ${x}^{2} - 5 x - 14 = 0$.
Factoring to get $\left(x - 7\right) \left(x + 2\right) = 0$

finds the solution set to be $\left\{7 , - 2\right\}$.
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, ${x}^{2} + 9 = {\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9$
Which leads to $0 = 6 x$ which has only one solution, $0$ which works in the original equation.