How do extraneous solutions arise from radical equations?

1 Answer
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: x1=4 and x=5, have exactly the same set of solutions. Namely: {5}.

Square both sides of x=5 to get the new equation: x2=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 x1=4 to get x22x+1=16
which is equivalent to x22x15=0.
and, rewriting the left, (x+3)(x5)=0.
So the solution set is {3,5}.
This time, it is 3, that is the extra solution.

Example2 : Extraneous solution.
Solve x=2+x+18
Subtracting 2 from both sides: x2=x+18

Squaring (!) gives x24x+4=x+18
This requires, x25x14=0.
Factoring to get (x7)(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 x2+9=x+3
Squaring (!) gives, x2+9=(x+3)2=x2+6x+9
Which leads to 0=6x which has only one solution, 0 which works in the original equation.