Lesson · Rational and Radical Expressions · Radicals

Solving Radical Equations

Isolate the radical and raise both sides to the index. The step is legal and it is not reversible, so it can produce numbers that solve the new equation and not the original. The check is not proofreading — it is the final step of the method.

Open the interactive lesson Practice

√(x + 7) = x − 5 x + 7 = x² − 10x + 25 squaring: legal, NOT reversible x² − 11x + 18 = 0 → x = 2 or x = 9 x = 9 checks; x = 2 is EXTRANEOUS

Two candidates, one solution

Core ideas

What this lesson establishes

One implication runs forwards and not backwards, and every difficulty in this lesson follows from that.

Squaring is a one-way implication

If a = b then a² = b² — always. The converse fails: a² = b² allows a = −b. So every solution of the original survives the squaring, but the squared equation may have picked up solutions of a = −b as well.

The extraneous root is not a mistake

It is the correct consequence of a legal step. No amount of care in the algebra prevents it, which is why the method has a checking step built in rather than a warning attached.

Check against the ORIGINAL equation

Substituting into a line partway down verifies your arithmetic from that line onwards, and that is not what is at issue. Only the original distinguishes a solution from an artefact of squaring.

The material

Solving equations that contain a radical

The method, the reason it needs a check, and the case of two radicals.

“Solving Rational Equations” met extraneous solutions when clearing denominators: multiplying by an expression that could be zero can create a root. This lesson meets them again from a different direction, and the cure is identical — substitute into the original.

Definition · Radical equation

A radical equation is one in which the variable appears under a radical sign. √(x + 7) = x − 5 is one; √7 · x = 5 is not, since the variable is outside.

Theorem · Raising both sides to a power

If a = b then aⁿ = bⁿ for every positive integer n. The converse holds for odd n, and fails for even n.

For even n, aⁿ = bⁿ gives a = ±b. Squaring an equation therefore solves a = b and a = −b at once, and the two solution sets come back mixed together.

Follow the key example. √(x + 7) = x − 5 squares to x + 7 = x² − 10x + 25, which rearranges to x² − 11x + 18 = 0 and factors as (x − 2)(x − 9) = 0. The candidates are 2 and 9.

At x = 9: the left side is √16 = 4 and the right side is 4. It solves the equation. At x = 2: the left side is √9 = 3 and the right side is −3. It does not — but it does solve √(x + 7) = −(x − 5), which is the other equation squaring silently included.

Example · Seen without any algebra

√(x + 1) = −3 has no solution, and no work is needed to say so: a principal square root is never negative. Squaring it produces x + 1 = 9 and the confident answer x = 8, which is entirely extraneous. The equation was unsolvable before the first step.

Two radicals

Isolate one radical on its own side before squaring, even if another remains. Squaring a side with two terms is where “Adding, Subtracting and Multiplying Radicals”'s middle term returns: (1 + √x)² = 1 + 2√x + x still contains a root, so nothing has been gained.

For √(x + 5) − √x = 1, move √x across to get √(x + 5) = 1 + √x. Squaring gives x + 5 = 1 + 2√x + x, so 4 = 2√x and √x = 2, hence x = 4. Check in the original: √9 − √4 = 3 − 2 = 1 ✓. Two squarings, one surviving solution.

For an odd index there is no extraneous-root problem, because cubing is reversible: a³ = b³ forces a = b for real numbers. The check is still worth doing as arithmetic insurance, but it is no longer part of the logic of the method.

Interactive

Radical equations

Square both sides, then check what squaring let in

Method

Solving a radical equation

Five steps, and the last one is not optional.

Isolate one radical on one side

Get a single radical alone before raising anything to a power. If two radicals are present, isolate the more complicated one first.

Raise both sides to the index

Square for a square root, cube for a cube root. Square the whole side, not term by term: (x − 5)² is x² − 10x + 25.

Repeat if a radical remains

Isolate and raise again. Each pass removes one radical, and a second squaring does not add a second layer of extraneous risk beyond what the final check handles.

Solve the resulting equation

It is usually linear or quadratic, and the factoring from Polynomials and Factoring handles it. These are candidates, not yet solutions.

Substitute every candidate into the ORIGINAL equation

Keep the ones that make it true and discard the rest. State the discarded ones as extraneous — an answer of "no solution" is a legitimate and complete result.

Which candidates survive

Score 0 / 0

Choose the answer you can justify.

One equation with an extraneous root, one with two radicals

In (a) the algebra is flawless and one of the two answers is still not a solution.

(a) √(x + 7) = x − 5 square both sides x + 7 = (x − 5)² = x² − 10x + 25 rearrange 0 = x² − 11x + 18 factor 0 = (x − 2)(x − 9) candidates x = 2, x = 9 CHECK in the original: x = 9: √(9 + 7) = √16 = 4 and 9 − 5 = 4 ✓ x = 2: √(2 + 7) = √9 = 3 and 2 − 5 = −3 ✗ solution set: { 9 } (b) √(x + 5) − √x = 1 isolate √(x + 5) = 1 + √x square x + 5 = 1 + 2√x + x ← middle term kept simplify 4 = 2√x, so √x = 2 square again x = 4 CHECK: √(4 + 5) − √4 = 3 − 2 = 1 ✓ solution set: { 4 }

In (a), x = 2 satisfies x² − 11x + 18 = 0 exactly. It is a real solution of the squared equation and not of the original, because squaring merged √(x + 7) = x − 5 with √(x + 7) = −(x − 5) — and 2 solves the second one.

There is a shortcut worth noticing in (a): the left side is a principal square root, so it is never negative, so x − 5 ≥ 0 and x ≥ 5 before any work is done. That condition alone rejects x = 2. Writing the domain restriction first is the same habit “Rational Expressions and Their Domains” asked for.

For a faded solve, take √(2x + 3) = x. The supplied sign condition is x ≥ 0, because a principal square root cannot equal a negative right side. Square, factor the resulting quadratic and check both candidates in the original. You should obtain 3 and −1, reject −1, and report {3}. Keeping −1 means the sign condition and the original-equation check were both applied only to the squared equation.

Common mistakes

Squaring term by term√(x + 7) = x − 5 does not become x + 7 = x² − 25. The whole side is squared, and (x − 5)² = x² − 10x + 25 keeps its middle term.

Squaring before isolatingWith √(x + 5) − √x = 1, squaring immediately leaves a cross term containing √(x² + 5x). Isolate one radical first and the second squaring is straightforward.

Treating the check as optional tidyingIt is a step of the method, not a courtesy. Without it the answer to example (a) is "2 and 9", which is wrong, and nothing earlier in the working reveals it.

Check your understanding

✓

Finish when you can say WHY a candidate failed, not merely that it did.

Given a radical equation you should be able to solve it, check both candidates in the original, and name the equation the extraneous root actually satisfies. You should also be able to spot, before starting, when a sign condition makes some candidates impossible.

Note: This lesson and “Solving Rational Equations” are the same lesson in two costumes. Clearing a denominator and squaring both sides are both legal, both irreversible, and both handled by the identical discipline: solve, then test every candidate against the equation you were given.