Rational and radical equations

Start with the equation as it was given

An algebraic rational expression is a ratio of polynomials with a nonzero denominator. Here rational equations include an unknown in a denominator, such as 3/(x − 1) = 2. Radical equations include an unknown under a root, such as √(x + 2) = x. An equation can contain both kinds of expression. Work over the real numbers in this lesson unless the problem gives a narrower domain.

The original domain consists of inputs for which every expression in the original equation is defined. A denominator must be nonzero, and a real square root must have a nonnegative radicand, the expression under the root. Record these conditions before cancelling anything. For 1/(x − 2) + 1/(x + 3) = 1, the original restrictions are x ≠ 2 and x ≠ −3. Neither restriction can disappear during later rewriting.

Being in the domain makes an input eligible for checking; it does not make the equality true. An excluded input is undefined in the original equation. A permitted input may instead give two defined but unequal sides. Keep these two reasons for rejection separate.

Cancel factors while keeping their nonzero conditions

Factor a denominator to find every forbidden input. Since x² − 9 = (x − 3)(x + 3), the expression 1/(x² − 9) requires x ≠ 3 and x ≠ −3. A denominator such as (x − 1)² excludes only x = 1, even though its factor is repeated. Check all original denominators, rather than only the ones visible after simplification.

Cancellation divides a numerator and denominator by a shared nonzero factor. Thus (x² − 9)/(x − 3) = x + 3 for x ≠ 3, because the numerator factors as (x − 3)(x + 3). At x = 3 the original fraction is 0/0, which is undefined; the value of x + 3 there does not repair it. You cannot cancel a term across addition: (x + 3)/x is 1 + 3/x for x ≠ 0, not 3.

A fraction equals zero exactly when its numerator is zero and its denominator is nonzero. For (x² − 4)/(x − 2) = 0, the numerator suggests x = −2 or x = 2. Only −2 is permitted: its fraction is 0/(−4) = 0, while 2 makes the original denominator zero.

Multiply the complete equation by a common denominator

After recording the original domain, factor the denominators and choose a least common denominator, or LCD, containing the needed factors with their highest powers. For denominators 2x and 3x², an LCD is 6x², with x ≠ 0. A common denominator need not be the smallest to work, but using the LCD usually reduces expansion.

Multiply both sides of the equation, including every term, by the LCD. For 1/x + 1/(x − 1) = 3, x ≠ 0 and x ≠ 1, so the LCD x(x − 1) gives (x − 1) + x = 3x(x − 1). Multiplying only the fractional terms and leaving the 3 unchanged would change the equation.

On the original domain the LCD is nonzero, so this multiplication can be reversed by division by the same LCD. The cleared equation and the original equation therefore have the same solutions within that domain. If you solve the cleared polynomial over all real numbers, filter its candidates through the recorded exclusions. A zero LCD cannot establish equivalence at an input where the original equation was undefined.

A simplified equation can have none, some or all permitted solutions

For (x² − 16)/(x − 4) = 8, first exclude x = 4. Cancelling the factor x − 4 on the permitted domain gives x + 4 = 8, whose only candidate is 4. That candidate is undefined in the original equation, so the original has no real solution. Clearing the denominator instead gives (x − 4)² = 0 and the same rejected candidate.

For (x² − 16)/(x − 4) = x + 4, the same cancellation gives x + 4 = x + 4 for x ≠ 4. Every permitted real input satisfies the equation, but 4 remains excluded. State the full answer as all real x except 4, rather than all real x. An identity after cancellation does not fill the missing original input.

A contradiction after valid clearing also gives no solution: 1/(x − 4) = 2/(x − 4) has domain x ≠ 4 and becomes 1 = 2. A remaining linear or quadratic equation may give a finite set of solutions. Inspect the resulting equation instead of assuming every rational equation has one answer.

A principal square root is always nonnegative

For F ≥ 0, √F denotes the principal, nonnegative square root. In particular, √9 = 3, and √(u²) = |u| for real u. The equation u² = 9 has solutions 3 and −3; the expression √9 itself has only the value 3. A radical equation does not automatically call for a ± sign.

For √F(x) = G(x), an original solution requires F(x) ≥ 0 and G(x) ≥ 0, as well as any other restrictions needed to define F and G. The first condition makes the square root real; the second follows from its nonnegative value. For √(x + 2) = x, the root is defined when x ≥ −2, but equality can hold only when x ≥ 0.

If a square root is the entire denominator, its radicand must be strictly positive. The expression 1/√(x − 1) needs x > 1: x = 1 gives a zero denominator and x < 1 gives no real square root. For a denominator containing additional terms, inspect the complete denominator: 1/(√x + 1) is defined for every x ≥ 0 because √x + 1 is then positive.

Isolate the radical, then square the whole other side

Use equivalent addition, subtraction and division by fixed nonzero coefficients to isolate a square root. From 2√(x + 5) + 3 = 9, subtract 3 and divide by 2 to get √(x + 5) = 3. Squaring then gives x + 5 = 9, hence x = 4. In the original equation, 2√9 + 3 = 9, so the candidate passes.

When the other side is a binomial, square the whole binomial. From √(x + 5) + 1 = x, isolate √(x + 5) = x − 1. Original definition requires x ≥ −5 and equality additionally requires x ≥ 1. Squaring gives x + 5 = (x − 1)² = x² − 2x + 1; the cross term −2x must remain.

The resulting equation is x² − 3x − 4 = 0, so its candidates are 4 and −1. Original checks give √9 + 1 = 4 at 4, but √4 + 1 = 3, not −1, at −1. The only original solution is 4. Squaring √(x + 5) + 1 directly would also be possible if expanded correctly, but it would leave the cross term 2√(x + 5) and require another isolation step.

Squaring preserves equality but can erase a sign difference

If two real expressions are equal, their squares are equal. The converse fails in general: 2 and −2 have equal squares but different values. Therefore solving F(x) = G(x)² after squaring √F(x) = G(x) supplies candidates until the original conditions and equality are checked.

When F and G are defined, √F(x) = G(x) is equivalent to the combined conditions F(x) = G(x)² and G(x) ≥ 0. The squared equality already makes F nonnegative. Retaining the original domain and this sign condition makes the squaring step reversible, because the nonnegative number whose square is G² is G itself.

A sign check can settle an equation before any squaring. The equation √(x + 5) = −2 has no real solution. Squaring would suggest x = −1, but its original left side is 2, not −2. If more than one square root remains, isolate a radical again before another squaring, retain every original radicand condition and check candidates in the original equation after the final step.

An equation can require both denominator and sign checks

Consider 1/√x = 1/(x − 2). Its original domain is x > 0 and x ≠ 2. Within that domain, multiplying both sides by the nonzero expression √x(x − 2) gives x − 2 = √x. Equality now requires x − 2 ≥ 0, so together with x ≠ 2 a solution must have x > 2.

Squaring the isolated equality gives (x − 2)² = x, or x² − 5x + 4 = 0. The polynomial candidates are 1 and 4. At x = 1 the original expressions are defined, but they give 1 = −1, which fails. At x = 4 they give 1/2 = 1/2, which passes. The complete original solution set is therefore {4}.

The two checks serve different purposes. An input such as 2 is excluded because an original denominator is zero. The candidate 1 is in the original domain but fails equality because squaring lost the sign difference. Neither the cleared equation nor the squared equation is a substitute for the original check.

Verify every candidate, then apply the problem's restrictions

For each candidate, first inspect every original denominator and radicand. If they are defined, evaluate both original sides with exact fractions or radicals. Reject a candidate if those exact values differ. Keep exact answers until these checks are complete; rounding a radical early may hide a difference or produce a small residual even for a valid root.

Original checking proves whether a particular candidate works. Completeness comes from the algebra: reversible clearing on the domain, a complete solution of the remaining linear or quadratic equation, and the necessary sign conditions for squared equations. Passing several chosen inputs does not prove an identity, and failing a finite list does not prove that no solution exists.

A problem may add conditions such as positive lengths, nonnegative elapsed times, integer counts or a stated interval. Apply those after the original algebraic checks and report units when needed. For example, (x − 2)(x − 3)/(x − 1) = 0 has real solutions 2 and 3. If the same model explicitly permits only x ≥ 5/2, only 3 answers that problem; 2 still satisfies the original real equation. Context restrictions and undefined original inputs are different reasons to remove a value.

Compare a complete solution set with an independent candidate

The explorer uses six fixed equations: a rational equation with two roots, a rational equation whose only cleared candidate is excluded, a cancellation identity with one excluded input, a square-root equation with a wrong-sign candidate, an equation with a square root in a denominator, and a square root equal to a positive constant. Choose the equation and candidate independently, then inspect the original domain, the transformed equation and the original evaluation.

The cancellation case is (x² − 1)/(x − 1) = x + 1: every real x except 1 works. The no-solution case is (x² − 4)/(x − 2) = 4: its only cleared candidate is 2, where the original is undefined. These cases show why a solution set may be an entire permitted domain or empty rather than a finite list of roots.

The finite candidate choices illustrate checks and cannot cover the entire real line. A failed candidate does not determine the full solution set, and a passing candidate need not be the only solution. Read the displayed full set together with the derivation that justifies it; then use the original evaluation to explain the chosen candidate's status.

Keep one chain from the original equation to the final set

A reliable solution records the original domain, factors where useful, clears denominators only on that domain, isolates any square root, retains sign conditions when squaring, solves the resulting equation completely and checks every candidate in the original. A factorisation is an identity; dividing by a variable factor requires that factor to be nonzero. Factoring a polynomial to use the zero-product rule can preserve a zero candidate that division would lose.

The fraction and square-root equations here reduce to linear or quadratic algebra from earlier lessons. No real solution means none within the chosen real domain. Complex numbers, higher-degree solution methods and function features such as asymptotes belong to later lessons; they are not needed to establish these solution sets. More detailed modelling with equations follows next, using the same domain and checking habits.

Worked examples

Try it yourself

1. State the original domain of (x + 2)/(x² − 4) + 1/(x + 2) = 0. Does cancelling x + 2 in the first fraction permit x = −2?

   Show the answer and reasoning

   The original domain is all real x except −2 and 2. Cancellation does not permit −2.

   The first denominator factors as (x − 2)(x + 2), excluding both values; the second also excludes −2. The first fraction can be written 1/(x − 2) only on the original domain. At −2 both original denominators are zero.

2. Solve 2/(x − 1) − 3/(x + 1) = 1 and check every candidate in the original equation.

   Show the answer and reasoning

   The complete real solution set is {−3, 2}; x = −1 and x = 1 are excluded.

   The LCD (x − 1)(x + 1) gives 2(x + 1) − 3(x − 1) = x² − 1, hence x² + x − 6 = (x + 3)(x − 2) = 0. Neither root is excluded. At −3 the original left side is −1/2 + 3/2 = 1; at 2 it is 2 − 1 = 1. Reversible clearing and complete factorisation establish that no other solutions remain.

3. Solve (x² − 25)/(x − 5) = 10. Explain why solving the simplified equation alone would give the wrong final answer.

   Show the answer and reasoning

   There is no real solution; the only simplified candidate is the excluded input 5.

   Record x ≠ 5. On that domain, cancel the factor x − 5 to get x + 5 = 10, whose only candidate is 5. In the original equation that input gives 0/0, so it is undefined. The simplification is equivalent only for x ≠ 5.

4. Find the complete solution set of (x² − 4)/(x − 2) = x + 2. Would checking only x = 0 and x = 3 establish your answer?

   Show the answer and reasoning

   Every real x except 2 satisfies the equation. Two checks alone do not establish the full set.

   Record x ≠ 2, then factor the numerator to (x − 2)(x + 2). Cancelling the nonzero factor on the domain proves the identity x + 2 = x + 2 for every permitted x. At 2 the original fraction is undefined. The factor identity, rather than a finite list of successful checks, establishes completeness.

5. Solve √(x + 6) = x. Which squared candidate is defined in the original equation but still fails it?

   Show the answer and reasoning

   The only solution is 3. The squared candidate −2 is defined but fails equality.

   The original domain is x ≥ −6, and equality requires x ≥ 0. Squaring gives x² − x − 6 = (x − 3)(x + 2) = 0, with candidates 3 and −2. Original checks give √9 = 3 at 3, but √4 = 2, not −2, at −2. The sign restriction and the complete factorisation leave only 3.

6. Solve √(2x + 3) + 1 = x, retaining exact answers and the middle term when squaring.

   Show the answer and reasoning

   The only real solution is 2 + √6.

   The original domain is x ≥ −3/2. Isolate √(2x + 3) = x − 1, which requires x ≥ 1. Square to obtain 2x + 3 = x² − 2x + 1, hence (x − 2)² = 6. Let r = √6 > 2: the candidates are 2 ± r. At 2 + r, the radicand is (1 + r)² and the original left side is 2 + r, so it passes. At 2 − r, the radicand is (r − 1)² and the original left side is r, unequal to 2 − r; it fails.

7. Solve 1/√(x − 2) = 1/3. Why must x = 2 be excluded even though its radicand is nonnegative?

   Show the answer and reasoning

   The original domain is x > 2, and the only solution is 11. At x = 2 the denominator is zero.

   A real nonzero square-root denominator requires x − 2 > 0. On that domain, multiplying by 3√(x − 2) gives √(x − 2) = 3. Squaring yields x − 2 = 9, so x = 11. The original check is 1/√9 = 1/3. The root at x = 2 would be real but unusable as a denominator.

What to carry forward

• Record the restrictions from every original denominator and radicand before cancellation, and keep excluded inputs excluded throughout.
• Cancel whole nonzero factors, not terms across addition; a fraction is zero only when its numerator is zero and its denominator is nonzero.
• Multiplying the complete equation by an LCD is reversible only on the original domain where that LCD is nonzero.
• A principal square root is nonnegative. The equation √F = G needs F ≥ 0 and G ≥ 0; a square root used as an entire denominator requires F > 0.
• Isolate a radical before squaring, square the whole other expression with its cross term, and treat unfiltered squared roots as candidates.
• Check every candidate in the original equation, then apply context and units. Use the algebra to justify completeness; a finite set of checks cannot do that alone.
• State the full solution set, whether it contains finitely many values, no values or every real input in the permitted domain.