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QUESTION

# 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.