How do I solve a rational equation and check for extraneous solutions?
Quick answer
First record every value that makes an original denominator zero. Multiply both sides by the least common denominator (LCD), solve the resulting equation, then check each candidate in the original equation. Reject restricted values: clearing denominators can produce a candidate that satisfies the transformed equation but makes the original equation undefined. For example, (x²−4)/(x−2)=0 gives candidates −2 and 2, but only −2 is valid.
The question, the details, the next step.
What order should I follow so I don’t accidentally accept a value that makes a denominator zero?
Use this sequence: 1. Write down the restrictions from the denominators in the equation as given. 2. Find the least common denominator and multiply every term by it. 3. Solve the resulting equation. 4. Check each candidate against the restrictions and in the original equation. The restriction check is essential: clearing denominators is an algebra step, not proof that every candidate works in the original equation. [S1]
For instance, if an original denominator is x−2, write x≠2 before doing any algebra. Keep that note visible until you have finished checking.
Can you work through an example where one candidate has to be rejected?
Solve:
(x²−4)/(x−2) = 0.
The denominator is zero at x=2, so the restriction is x≠2. The LCD is x−2. Multiply both sides by it, keeping in mind that this step applies only to values allowed by the restriction:
x²−4 = 0.
Factor and solve:
(x−2)(x+2)=0,
so the candidates are x=2 and x=−2. Now return to the original equation. At x=2, the denominator x−2 is zero, so the original left-hand side is undefined; reject 2. At x=−2, the numerator is 4−4=0 and the denominator is −4, so the fraction equals 0. Thus the solution is x=−2.
Why does multiplying by the LCD not guarantee that all the answers to the new equation work?
The original rational equation is defined only where its denominators are nonzero. Multiplying by the LCD turns it into an equation without fractions, and that new equation may have a root at a value excluded from the original domain. In the example, x=2 makes x²−4 equal zero, so it solves the cleared equation, but it makes the original denominator zero. That candidate is extraneous because the original equation is undefined at that value. [S1]
Carry the restrictions alongside your work, then substitute every candidate into the original equation. The check at −2 confirms that the original fraction equals 0.
What can I write down as a compact check for my homework?
Use this copyable layout and fill in each blank:
Equation: [original rational equation] Restrictions from original denominators: [variable cannot equal …] LCD: [least common denominator] After multiplying and simplifying: [equation without fractions] Candidates: [values] Original-equation checks: [substitute each candidate; mark undefined or show both sides] Answer: [candidates that satisfy the original equation]
For this example, the filled-in result is: restriction x≠2; LCD x−2; cleared equation x²−4=0; candidates 2 and −2; reject 2 because the original expression is undefined there; keep −2 because the original fraction equals 0. [S1]
A practical checklist
List values that make any original denominator zero before multiplying.
Use the LCD to clear every fraction in every term.
Solve the resulting equation and record all candidates.
Reject candidates that violate an original restriction.
Substitute each remaining candidate into the original equation and verify both sides.
A few more useful answers
Is a restricted candidate always an extraneous solution?
If algebra produces it as a candidate, yes: it must be rejected because the original equation is undefined there. It is not a solution of the original equation.
Can I check candidates only in the equation after clearing denominators?
No. That equation may include a candidate excluded by an original denominator. Check the original equation and its restrictions.
What if every candidate is rejected?
Then none of the candidates satisfies the original equation, so it has no solution among them. State that result and show the restriction and substitution checks.
Sources
References used for this guide.
- [S1] OpenStax Intermediate Algebra: Solve Rational Equations openstax.org · Checked