Solving Radical Equations

To solve an equation with a square root:

1. Isolate the radical on one side.
2. Square both sides.
3. Solve the resulting equation.
4. Check every answer in the *original* equation.

Step 4 is mandatory. Squaring can create false solutions, because squaring destroys sign information: −3 and 3 both square to 9.

For √(x + 6) = x: squaring gives x + 6 = x², so x² − x − 6 = 0, giving x = 3 or x = −2. Checking: √9 = 3 ✓, but √4 = 2 ≠ −2 ✗. Only x = 3 works.

If there are two radicals, isolate one, square, then repeat.

Squaring is not injective on ℝ, so A = B ⟹ A² = B² holds but the converse does not. The squared equation's solution set contains the original's and may be strictly larger, producing extraneous roots. Checking in the original equation is part of the method.

Isolation before squaring is required, since (a + √b)² = a² + 2a√b + b retains a radical, defeating the purpose. With two radicals, isolate and square repeatedly; each squaring reduces the count by one.

An equivalent framing that avoids surprises: √(f(x)) = g(x) is equivalent to f(x) = g(x)² *together with* the condition g(x) ≥ 0, because the principal square root is non-negative. In the example above, that condition immediately rules out x = −2 without substitution.

Equations with odd-index radicals do not produce extraneous solutions from cubing, since cubing is injective — the phenomenon is specific to even powers.