Radical Equations and Extraneous Solutions
A radical equation is an equation in which the unknown appears under a root sign, and solving it requires isolating the root and raising both sides to a power. In brief: to solve √(x + 3) = x − 3, square both sides, solve the resulting quadratic and then check each answer in the original equation, because squaring can create extraneous solutions. Here x = 6 works and x = 1 does not.
The check at the end is not optional. Raising both sides of an equation to an even power can turn a false statement into a true one, so the new equation may have more solutions than the original. This guide shows the method, explains why extraneous solutions occur and solves several worked examples with one or two radicals.
How do you solve a radical equation step by step?
- Find the domain: the radicand of any even root must be at least 0.
- Isolate one radical on one side of the equation.
- Raise both sides to the power that matches the index: square for a square root, cube for a cube root.
- Solve the resulting equation. If a radical remains, isolate it and repeat.
- Check each candidate in the original equation. Reject any that fail.
Step 3 is based on the idea that if a = b, then an = bn. The reverse is not always true, which is the source of extraneous solutions.
What is an extraneous solution and why does it appear?
An extraneous solution is a value that satisfies the transformed equation but not the original one. Consider x = 3. Squaring both sides gives x2 = 9, which has the solutions 3 and −3. But −3 does not satisfy x = 3. Squaring loses information about the sign.
For the same reason, in an equation like √A = B, the right side B must be nonnegative, since the left side is never negative. Any candidate that makes B negative is automatically extraneous. This quick test often rejects wrong answers before the full check.
Worked examples
Example 1: √(2x + 1) = 5
The radical is already isolated. Square both sides: 2x + 1 = 25, so x = 12. Check: √(2 · 12 + 1) = √25 = 5. ✓ The solution is x = 12.
Example 2: √(x + 3) = x − 3
The domain requires x ≥ −3. Since the left side is nonnegative, we also need x − 3 ≥ 0, so x ≥ 3. Squaring gives x + 3 = x2 − 6x + 9, so x2 − 7x + 6 = 0, which factors as (x − 1)(x − 6) = 0. The candidates are 1 and 6.
Check x = 1: √4 = 2, but x − 3 = −2. This fails, so x = 1 is extraneous. Check x = 6: √9 = 3 and 6 − 3 = 3. ✓ The only solution is x = 6.
Example 3: ∛(x − 5) = 2
Cube both sides: x − 5 = 8, so x = 13. For odd roots, cubing is reversible, so no extraneous solutions arise. Check: ∛8 = 2 ✓.
Example 4: x + 2 = √(4x + 5)
Squaring gives x2 + 4x + 4 = 4x + 5, so x2 = 1 and x = ±1. Check x = 1: left side 3, right side √9 = 3 ✓. Check x = −1: left side 1, right side √1 = 1 ✓. Both solutions are valid, which shows that not every equation has an extraneous solution.
How do you solve an equation with two radicals?
Isolate one radical, square, then isolate the remaining radical and square again.
Example: √(x + 5) − √(x) = 1
Isolate: √(x + 5) = √x + 1. Square both sides: x + 5 = x + 2√x + 1, so 4 = 2√x and √x = 2, giving x = 4. Check: √9 − √4 = 3 − 2 = 1 ✓.
Note that you must expand (√x + 1)2 fully as x + 2√x + 1, not as x + 1. The middle term is easy to forget.
Quick reference table
| Equation type | Operation | Extraneous solutions? |
|---|---|---|
| Square root = number | Square both sides | Possible |
| Square root = expression | Square both sides | Possible |
| Cube root = expression | Cube both sides | No |
| Two square roots | Square twice | Possible |
| Fractional exponent | Raise to the reciprocal power | Possible for even roots |
Where does this technique come from?
Solving equations with roots goes back to ancient problems about geometric lengths, and rules for removing roots by squaring appear in medieval Arabic algebra and in European texts of the sixteenth century. The systematic warning about extra roots introduced by squaring became standard in algebra teaching as the understanding of equivalent transformations improved.
Where are radical equations used today?
The distance formula contains a square root, so finding a point at a given distance from another point leads to a radical equation. The fall time t = √(2h/g) can be solved for h by squaring. Equations for the speed of a pendulum, the escape velocity or the length of a ladder leaning against a wall all use the same approach. In each case the check step makes sure that the answer has a physical meaning, for example a nonnegative length.
Which mistakes are common?
Skipping the check
This is the most expensive error. Always substitute the candidates into the original equation.
Squaring a sum term by term
(√x + 1)2 = x + 2√x + 1, not x + 1. Use the binomial square.
Squaring before isolating the radical
In √(x + 1) + 2 = x, first move the 2 to get √(x + 1) = x − 2, and only then square.
Rejecting a valid solution
A solution that makes the radicand zero is fine, as in √(x − 4) = 0 with x = 4. Check carefully before discarding an answer.