Solve simple rational and radical equations in one variable and explain how extraneous solutions arise

Learning goal: Solve simple rational and radical equations in one variable by rewriting them as polynomial equations, check every candidate in the original equation, and give examples showing how extraneous solutions may arise from squaring both sides or from multiplying by an expression that can be zero.

Two families of equations. A radical equation is an equation in which the variable appears under a root symbol, such as √(2x + 7) = x + 2 or ∛(3x − 1) = 2. A rational equation is an equation in which the variable appears in a denominator, such as x/(x − 3) = 3/(x − 3) + 2. Both families are solved by the same strategy: perform an operation on both sides that removes the root or the denominator, solve the polynomial equation that results, and then test each candidate in the original equation. The last step is not optional politeness. It is part of the method, because the operations that remove roots and denominators can create solutions that the original equation never had.

Why solving is a chain of reasoning. When we solve an equation we are claiming: if x satisfies the original equation, then x satisfies the next equation, and so on down to the last line. Each line is a consequence of the one before it. Some operations are reversible: adding 5 to both sides can be undone by subtracting 5, and multiplying both sides by a nonzero constant can be undone by dividing. For reversible operations the two equations have exactly the same solutions. But squaring both sides is not reversible, and multiplying both sides by an expression that might be zero is not reversible either. After such a step the new equation keeps every solution of the old one and may gain new ones. A candidate that appears in the final line but fails the original equation is called an extraneous solution: it is a solution of the derived equation, not of the equation we were asked to solve.

How squaring creates extraneous solutions. The numbers 3 and −3 are different, yet 3² and (−3)² are both 9. So the statement a = b and the statement a² = b² are not equivalent: a² = b² is also true when a = −b. Consider √(2x + 7) = x + 2. Squaring gives 2x + 7 = x² + 4x + 4, so x² + 2x − 3 = 0, which factors as (x + 3)(x − 1) = 0, giving the candidates x = 1 and x = −3. Test x = 1: the left side is √(2 + 7) = √9 = 3, and the right side is 1 + 2 = 3, so x = 1 is a solution. Test x = −3: the left side is √(−6 + 7) = √1 = 1, and the right side is −3 + 2 = −1. The two sides are 1 and −1: their squares agree, which is exactly why squaring let x = −3 through, but the sides themselves do not. The solution set is {1}, and x = −3 is extraneous. Notice the general shape of this example: the square root symbol means the non-negative root, so √(anything) = x + 2 can only hold when x + 2 ≥ 0, that is x ≥ −2. The candidate −3 breaks that condition, and squaring erased the condition.

How multiplying by a denominator creates extraneous solutions. A rational equation has restrictions before you start: any value that makes a denominator zero is excluded from the domain, because division by zero is undefined. Write the restrictions down first. For x/(x − 3) = 3/(x − 3) + 2 the restriction is x ≠ 3. Multiplying both sides by (x − 3) gives x = 3 + 2(x − 3) = 2x − 3, so x = 3. That is the one value we excluded. Substituting x = 3 into the original equation produces 3/0 on both sides, which is meaningless, so the equation has no solution at all. The multiplication step was the culprit: clearing the denominator gives an equation that agrees with the original only on the original domain, x ≠ 3; the polynomial equation knows nothing about that restriction, so it can produce the excluded value.

A rational equation with a genuine and an extraneous candidate — a full model. Solve x/(x + 2) + 1/(x − 2) = 8/(x² − 4). Factor the last denominator: x² − 4 = (x + 2)(x − 2). Restrictions: x ≠ −2 and x ≠ 2. The least common denominator is (x + 2)(x − 2). Multiply every term by it: x(x − 2) + (x + 2) = 8. Expand: x² − 2x + x + 2 = 8, so x² − x − 6 = 0, which factors as (x − 3)(x + 2) = 0. Candidates: x = 3 and x = −2. The candidate x = −2 is excluded by the restrictions, so it is extraneous. Check x = 3 in the original equation: 3/5 + 1/1 = 3/5 + 5/5 = 8/5, and the right side is 8/(9 − 4) = 8/5. The sides agree, so the solution set is {3}. Notice that the extraneous candidate is not a mistake in arithmetic; it is a value that the multiplication step manufactured.

Equations with two radicals. When two square roots appear, isolate one of them before squaring, and expect to square twice if necessary. Solve √(x + 5) − √x = 1. Isolate: √(x + 5) = 1 + √x. Square: x + 5 = 1 + 2√x + x, so 4 = 2√x and √x = 2, giving x = 4. Check: √9 − √4 = 3 − 2 = 1. It works. Isolating first is the simpler route: squaring √(x + 5) − √x directly produces a cross term −2√(x(x + 5)) that still contains both radicals, so that radical must then be isolated and squared again (it does still lead to x = 4, with more work).

Roots with odd index do not create extraneous solutions by themselves. Cubing is reversible, because every real number has exactly one real cube root: if a³ = b³ then a = b. So ∛(3x − 1) = 2 becomes 3x − 1 = 8, x = 3, and the check ∛8 = 2 confirms it, with no possibility of an extraneous candidate arising from the cubing step. It is the even-index roots, square roots and fourth roots, that hide sign information.

Equations with no solution at all. The equation √(x − 1) = −2 has no solution, because a principal square root is never negative. Squaring produces x − 1 = 4, x = 5, and the check gives √4 = 2, not −2. Every candidate is extraneous. Recognizing this before squaring saves work, but the check catches it either way.

The misconception and why it is wrong. Students often believe that if the algebra was done correctly, the answer must be right, and that checking is only for catching arithmetic slips. That is wrong for these equations. The algebra in every example above was correct, and extraneous candidates appeared anyway, because squaring and multiplying by variable expressions are one-way operations. A complete solution to a simple rational or radical equation always has three parts: the restrictions or sign conditions, the derived polynomial equation, and the check of each candidate against the original.

Worked radical equation with an extraneous candidate: Solve √(x + 7) = x + 1. Because a square root is non-negative, any solution needs x + 1 ≥ 0. Square both sides: x + 7 = x² + 2x + 1, so x² + x − 6 = 0, which factors as (x + 3)(x − 2) = 0. Candidates: x = 2 and x = −3. Check x = 2: √9 = 3 and 2 + 1 = 3, so it is a solution. Check x = −3: √4 = 2 but −3 + 1 = −2; the sides are opposites, which is exactly the situation squaring cannot tell apart. The solution set is {2}, and −3 is extraneous.

Worked rational equation with two genuine solutions: Solve 1/x + 1/(x + 2) = 3/4. Restrictions: x ≠ 0 and x ≠ −2. Multiply by the least common denominator 4x(x + 2): 4(x + 2) + 4x = 3x(x + 2), so 8x + 8 = 3x² + 6x, giving 3x² − 2x − 8 = 0, which factors as (3x + 4)(x − 2) = 0. Candidates: x = 2 and x = −4/3. Neither is excluded. Check x = 2: 1/2 + 1/4 = 3/4. Check x = −4/3: 1/(−4/3) = −3/4, and x + 2 = 2/3 so 1/(2/3) = 3/2; then −3/4 + 3/2 = −3/4 + 6/4 = 3/4. Both candidates survive, so the solution set is {2, −4/3}. Not every problem has an extraneous candidate; the check is what tells you.

Worked rational equation whose only candidate is extraneous: Solve (x + 2)/(x − 4) = 6/(x − 4). Restriction: x ≠ 4. Multiplying by (x − 4) gives x + 2 = 6, so x = 4. That is the excluded value, so the equation has no solution. A shorter argument reaches the same place: two fractions with the same nonzero denominator are equal only when their numerators are equal, and x + 2 = 6 forces x = 4, where the denominator is zero.

Worked cube-root equation: Solve ∛(2x + 3) = −1. Cube both sides: 2x + 3 = −1, so x = −2. Check: ∛(−4 + 3) = ∛(−1) = −1. Cubing is reversible, so this check could not have failed for a sign reason; it is still done, because it guards against arithmetic errors.

Reasoning record: Make a three-column table headed operation, reversible?, and why. Fill it for these operations: add a constant to both sides; multiply both sides by 5; multiply both sides by (x − 3); square both sides; cube both sides; take the square root of both sides of an equation of the form a² = b². For each non-reversible operation, write one specific equation in one variable where that operation produces an extraneous candidate, and show the check that exposes it.

Independent performance: Write and solve six original equations: two radical equations with square roots, one with a cube root, and three rational equations, at least one of which has two denominators that must be combined into a least common denominator. Design at least two of the six so that an extraneous solution arises, and at least one so that every candidate is genuine. For each equation, state the restrictions or sign conditions first, show the derived polynomial equation, list every candidate, and show the substitution check of each candidate in the original equation with both sides evaluated. Finish with a written paragraph that gives examples showing how extraneous solutions may arise, naming the two operations responsible and explaining, using the numbers from your own equations, what information each operation throws away. Success criteria: every excluded value is stated before the denominators are cleared; every candidate is checked in the original equation, not in a derived one; and the paragraph explains the mechanism rather than only reporting that a check failed.