Radical Equations and Extraneous Solutions

A square root only outputs non-negative numbers. So √(x + 6) = x only makes sense if x is non-negative too. But when you square both sides to clear the radical, you get x + 6 = x², an equation that no longer remembers that restriction. Squaring treats a number and its negative the same way: (-2)² and 2² both equal 4. That means the squared equation can pick up a solution where the original left side (always non-negative) would never have matched the right side. The algebra didn't lie to you, it just stopped tracking a piece of information you needed.

Solving √(x + 6) = x: square both sides to get x + 6 = x², rearrange to x² - x - 6 = 0, and factor to (x-3)(x+2) = 0, giving x = 3 or x = -2. Check both in the original equation. x = 3 gives √9 = 3, true. x = -2 gives √4 = -2, which is false since √4 = 2, not -2. So x = -2 is extraneous and gets thrown out, leaving x = 3 as the only real solution. The squaring step is irreversible, which is exactly why the check back into the original equation isn't optional busywork, it's the step that actually finishes the problem.

This sits alongside other equation types in the algebra toolkit where the same habit of checking answers matters, like rational equations where a solution can make a denominator zero. The common thread is that certain algebraic moves, squaring or multiplying by a variable expression, can change the solution set. Treat the squared or cleared equation as a list of candidates, not a list of answers.