HSA.REI.A.2: Solving Rational and Radical Equations and Extraneous Solutions
In plain English: HSA.REI.A.2 is the Common Core algebra standard that asks students to solve simple rational and radical equations in one variable and to give examples showing how extraneous solutions arise. Clearing denominators or squaring both sides can produce values that fail the original equation, so every candidate must be checked there. It is usually taught in Algebra II.
Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Understand solving equations as a process of reasoning and explain the reasoning Also written as HSA-REI.A.2 or A-REI.2 · Official standard
Students solve simple rational equations by multiplying both sides by a common denominator, and simple radical equations by isolating the radical and raising both sides to a power. The central idea is why these methods can produce extraneous solutions: values that the new equation accepts but the original equation does not. Multiplying by an expression that equals zero for some x, or squaring both sides, is a step that cannot always be undone, so the new equation can have more solutions than the original.
Students learn to list excluded values before solving a rational equation, to check every candidate in the original equation, and to build their own examples showing how an extraneous solution appears. They also see that cubing both sides does not create extraneous solutions, and that a context can rule out a valid algebraic solution (a negative speed, for example), which is a different issue from an extraneous one.
Learning Objectives
By the end of this lesson, students will be able to:
Solve simple rational equations in one variable by identifying excluded values and multiplying by a common denominator
Solve simple radical equations in one variable by isolating the radical and raising both sides to the appropriate power
Check every candidate solution in the original equation and identify extraneous solutions
Explain why multiplying by a variable expression or squaring both sides can create extraneous solutions, and give an example of each
Prior Knowledge Required
Students should already be comfortable with:
Justifying the steps in solving an equation and checking solutions HSA.REI.A.1
Solving equations of the form x² = p and x³ = p with square and cube roots 8.EE.A.2
Solving quadratic equations by factoring HSA.REI.B.4
Finding a least common denominator and multiplying binomials
Post the following and have students work alone for 3 minutes, then compare with a partner:
Warm-Up Prompt
"Solve x = 3. Now square both sides to get x² = 9 and solve that. Do the two equations have the same solutions? What went wrong, if anything?"
Students should see that x² = 9 has solutions 3 and -3, but only 3 solves x = 3. Squaring kept the true solution and added a new one. Ask: "If a = b, is a² = b² always true? If a² = b², is a = b always true?" Record the answer (yes; no, a could equal -b). This idea drives the whole lesson.
Direct Instruction20-25 minutes
Present the two solving routines side by side:
Rational equations: list the excluded values (any x that makes a denominator 0), multiply every term on both sides by the least common denominator, and solve the resulting equation.
Radical equations: isolate the radical on one side, raise both sides to the power that matches the index (square for √, cube for ∛), and solve the resulting equation.
Check every candidate in the original equation: reject any value that is excluded or that makes the original equation false. Those values are extraneous.
Explain the source: multiplying by an expression that can be 0, or squaring both sides, can add solutions. Cubing both sides cannot.
Rational, no extraneous solution
Solve 3/x + 1/2 = 5/(2x). Excluded value: x = 0. Multiply every term by 2x.
Equation: 6 + x = 5, so x = -1. Check: 3/(-1) + 1/2 = -5/2 and 5/(-2) = -5/2.
Rational, one extraneous solution
Solve 1 + 2/(x - 3) = 12/(x² - 9). Excluded values: x = 3 and x = -3. Multiply by (x - 3)(x + 3).
Equation: x² - 9 + 2(x + 3) = 12 → x² + 2x - 15 = 0 → x = -5 or x = 3. Reject x = 3 (excluded). Solution: x = -5.
Radical, one extraneous solution
Solve √(x + 7) = x - 5. The radical is already isolated, so square both sides.
Equation: x + 7 = x² - 10x + 25 → x² - 11x + 18 = 0 → x = 9 or x = 2. Check: √16 = 4 = 9 - 5, but √9 = 3 ≠ 2 - 5. Solution: x = 9.
Solve ∛(x - 2) = -3. Cube both sides. Every real number has exactly one real cube root, so cubing cannot add solutions.
Equation: x - 2 = -27 → x = -25. Check: ∛(-27) = -3.
After Example 3, show Diagram 1. The graphs of y = √(x + 7) and y = x - 5 meet only at x = 9. The extraneous value x = 2 is where the line meets y = -√(x + 7). Squaring both sides turned the equation into one that accepts both branches, which is exactly why x = 2 appeared.
Guided Practice15 minutes
Pairs solve two equations and classify every candidate as a solution or extraneous: 2x/(x - 1) = 2/(x - 1) + 1 (clearing gives x = 1, which is excluded, so there is no solution) and √(x + 2) = x (candidates 2 and -1; only x = 2 checks, because √1 = 1, not -1). Circulate and listen for students who stop after solving the quadratic without checking, or who square √x + 2 as x + 4. Debrief by asking each pair to state which step produced the extraneous value.
Independent Practice10-15 minutes
Students solve three problems alone: 5/(x + 1) = 2 (x = 3/2), √(3x + 1) = x - 1 (candidates 5 and 0; only x = 5 checks), and ∛(4x) = 2 (x = 2). For each problem students write one sentence: "The step that could have created an extraneous solution was..." or "No step could create an extraneous solution because..."
Closure5-10 minutes
Exit ticket: "Write your own radical equation that has an extraneous solution. Solve it, show which value is extraneous, and explain which step created it." A quick example to share afterward: √x = x - 2 gives x² - 5x + 4 = 0, so x = 4 or x = 1, and x = 1 is extraneous because √1 = 1 but 1 - 2 = -1.
Differentiation Strategies
For Struggling Students
Give a checklist card: excluded values, clear or isolate, solve, check each candidate in the original
Start with equations whose radical is already isolated, such as √(x + 3) = 4, before equations that need isolating first
Have students record every candidate in a table with columns for the left side, the right side and "true or false" in the original equation
For Advanced Students
Ask for a rational equation with two candidates where both are extraneous, and explain why the equation has no solution
Solve equations with two radicals, such as √(x + 5) = 1 + √x (x = 4), which require isolating and squaring twice
Compare an extraneous solution with a solution that is valid algebraically but rejected by a context (for example, a negative speed)
Assessment Guidance
What to Look For
The standard has two parts: solving, and giving examples of how extraneous solutions arise. Check that students test every candidate in the original equation, not in a later step, and that they can name the step responsible (multiplying by an expression that can be 0, or squaring). A student who lists x = 2 and x = 9 for √(x + 7) = x - 5 has solved the squared equation, not the original one.
02
Classroom Activities
3 Activities
1
Keep or Reject?
15 minPairs
Each pair gets six cards. Each card shows an equation and the candidate values produced by a correct solving process. Pairs substitute each candidate into the original equation and sort it into a "solution" pile or an "extraneous" pile.
Card Set
√(x + 7) = x - 5; candidates 9 and 2 (keep 9, reject 2)
√(x - 1) = x - 3; candidates 5 and 2 (keep 5, reject 2)
1/(x - 2) + 1/(x + 2) = 4/(x² - 4); candidate 2 only (reject, so no solution)
Procedure
Pairs substitute each candidate into the original equation and record the value of each side
For every rejected value, pairs write whether it failed because it is excluded (a denominator is 0) or because the two sides are not equal
Debrief: which cards had no extraneous solutions, and what do they have in common?
Modification for Distance Learning
Put the cards on a shared slide with two columns labeled "Solution" and "Extraneous." Pairs drag each candidate into place and type the substitution beside it.
2
Build an Extraneous Solution
20 minGroups of 3
Groups reverse the process. They start from a value they want to be extraneous and design an equation that produces it, which is exactly what the standard asks students to do: give examples showing how extraneous solutions arise.
Tasks
Radical: choose a line y = x + k and the curve y = √x. Squaring √x = x + k gives a quadratic; find a k for which exactly one root is extraneous (k = -2 gives candidates 4 and 1, and 1 is extraneous)
Rational: write an equation with (x - 4) in the denominators whose only candidate is x = 4 (for example, x/(x - 4) = 4/(x - 4))
Cube root: try to build a cube-root equation with an extraneous solution, and explain why it is not possible
Discussion Questions
Which step in each solving process can add a solution, and why?
Why can checking in the squared equation never catch an extraneous solution?
Can multiplying both sides by a nonzero number ever create an extraneous solution?
3
Rate Problems With Rational Equations
20 minPairs
Pairs set up and solve rational equations from rate and work contexts, then decide whether each algebraic solution makes sense in the situation.
Problems
A kayaker paddles 12 miles. If she paddled 2 mph faster, the trip would take 1 hour less. Solve 12/r - 12/(r + 2) = 1 (r = 4 mph; the other root, r = -6, is not a meaningful speed)
One hose fills a pool in 6 hours and a second hose fills it in 3 hours. Solve 1/6 + 1/3 = 1/t for the time together (t = 2 hours)
A printer finishes a job in 5 hours alone. With a second printer the job takes 3 hours. Solve 1/5 + 1/p = 1/3 for the second printer alone (p = 7.5 hours)
Requirements
Define the variable with units and write the equation before solving
Check each solution in the original equation
Explain whether a rejected value was extraneous (fails the equation) or valid but unrealistic (fails the context)
Gallery Walk Variation
Each pair writes its own rate problem on chart paper, with a solution key on the back. Pairs rotate, solve one other problem, and flip the paper to check.
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Diagrams & Visual Aids
2 diagrams
Diagram 1: Where an Extraneous Solution Comes From
Graphs drawn to scale (1 grid unit = 1). The only intersection of y = √(x + 7) and y = x - 5 is (9, 4), so x = 9 is the only solution. Squaring both sides also accepts the intersection with the dashed curve y = -√(x + 7) at (2, -3), which is why x = 2 appears as a candidate and fails the check.
Diagram 2: Checking Every Candidate
The two tests every candidate must pass. For rational equations, test for excluded values first. For radical equations, the key test is substituting into the original equation, since a principal square root is never negative.
04
Homework Assignment
~30 min
HSA.REI.A.2 Homework: Rational and Radical Equations
Directions: Solve each equation. List any excluded values before solving a rational equation. Check every candidate in the original equation and label each one as a solution or extraneous. When a value is extraneous, name the step that produced it.
One pump can empty a flooded basement in 6 hours. A larger pump can empty it in 4 hours. Write and solve a rational equation to find how long the job takes with both pumps running together.
Part 2: Radical Equations (Problems 4-5)
Solve √(3x + 4) = x.
Solve √(x + 1) + 7 = 3. Explain how you could tell there is no solution before squaring.
Part 3: Explain and Create (Problem 6)
(a) Write a rational equation whose only candidate solution is x = 6 and show that 6 is extraneous. (b) A student solves √(2x + 7) = x + 2 by squaring and reports x = 1 and x = -3. Check both candidates in the original equation, identify the extraneous one, and explain which step created it.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Solving Process
Correct excluded values and correct clearing or squaring
Correct method with an algebra error
Method missing or incorrect
Checking
Every candidate checked in the original equation
Some candidates checked, or checked in a later equation
No checking
Extraneous Solutions
Correctly labeled, with the step that caused them
Correctly labeled, no explanation
Not identified
Examples and Context
Own examples work and context answers are interpreted
Example has a minor flaw
Missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Solve 6/x = 3/(x - 2).
Answer: C
Cross-multiply: 6(x - 2) = 3x, so 6x - 12 = 3x and x = 4. Check: 6/4 = 1.5 and 3/2 = 1.5. Choice A comes from distributing incorrectly (6x - 2 = 3x). Choice D is an excluded value, since it makes x - 2 = 0.
Question 2 of 20 · Multiple Choice
Solve √(x + 5) = 4.
Answer: B
Square both sides: x + 5 = 16, so x = 11. Check: √16 = 4. Choice A comes from not squaring (x + 5 = 4), and choice D comes from doubling instead of squaring (x + 5 = 8).
Question 3 of 20 · Multiple Choice
When √(x + 6) = x is solved by squaring, the candidates are 3 and -2. Which value is extraneous?
Answer: A
For x = -2: √4 = 2, but the right side is -2, so it fails. For x = 3: √9 = 3, which checks. Choice D is the result of checking in the squared equation x + 6 = x², which both values satisfy.
Question 4 of 20 · Multiple Choice
What are the excluded values for 3/(x - 4) + 1/(x + 1) = 2?
Answer: D
A value is excluded if it makes a denominator 0: x - 4 = 0 gives x = 4 and x + 1 = 0 gives x = -1. Choice A reverses the signs, a common error when reading the factors.
Question 5 of 20 · Multiple Choice
Why can squaring both sides of an equation create an extraneous solution?
Answer: B
Squaring keeps every true solution, since a = b implies a² = b². The reverse fails: (-3)² = 3², but -3 ≠ 3. So the squared equation can have extra solutions where the original sides are opposites. Choice D is too strong; squaring is valid, it just requires a check.
Question 6 of 20 · Multiple Choice
Solve (x + 3)/(x - 1) = 2.
Answer: C
Multiply both sides by x - 1: x + 3 = 2x - 2, so x = 5. Check: 8/4 = 2. Choice A comes from not distributing the 2 (x + 3 = 2x - 1). Choice B is excluded, since it makes the denominator 0.
Question 7 of 20 · Multiple Choice
Solve 3x/(x - 2) = 6/(x - 2) + 2.
Answer: C
Multiply by x - 2: 3x = 6 + 2(x - 2), so 3x = 2x + 2 and x = 2. But x = 2 makes the denominators 0, so it is extraneous and the equation has no solution. Choice A is the candidate that was never checked.
Question 8 of 20 · Multiple Choice
What is the solution set of √(2x + 3) = x?
Answer: B
Squaring gives 2x + 3 = x², so x² - 2x - 3 = 0 and x = 3 or x = -1. Check: √9 = 3 works, but √1 = 1 ≠ -1. Choice A lists the solutions of the squared equation without checking.
Question 9 of 20 · Multiple Choice
Solve ∛(2x + 1) = 3.
Answer: D
Cube both sides: 2x + 1 = 27, so x = 13. Check: ∛27 = 3. Choice A comes from squaring instead of cubing (2x + 1 = 9). Cubing cannot create extraneous solutions, because every real number has exactly one real cube root.
Question 10 of 20 · Multiple Choice
Solve x + 8/x = 6.
Answer: A
Multiply by x (x ≠ 0): x² + 8 = 6x, so x² - 6x + 8 = 0 and x = 2 or x = 4. Check: 2 + 4 = 6 and 4 + 2 = 6. Both are solutions; multiplying by x only creates an extraneous solution if a candidate is 0. Choice B comes from a sign error when factoring.
Question 11 of 20 · Multiple Choice
Solve √(x + 4) = x - 2.
Answer: B
Square: x + 4 = x² - 4x + 4, so x² - 5x = 0 and x = 0 or x = 5. Check: √9 = 3 = 5 - 2 works, but √4 = 2 ≠ 0 - 2 = -2. Choice C skips the check.
Question 12 of 20 · Multiple Choice
A student solves √x + 2 = 5 by squaring both sides and writing x + 4 = 25. What is the error, and what is the correct solution?
Answer: D
Subtract 2 first: √x = 3, so x = 9. Check: √9 + 2 = 5. Squaring the left side as it stands gives x + 4√x + 4, not x + 4, so choice A uses an invalid step and gets 21, which does not check (√21 + 2 ≈ 6.58).
Question 13 of 20 · Multiple Choice
Which equation produces an extraneous solution when it is solved by squaring or cubing both sides?
Answer: C
Squaring √(x + 1) = -2 gives x + 1 = 4, so x = 3. But √4 = 2, not -2, so x = 3 is extraneous and there is no solution. A principal square root cannot be negative. Choice D is a trap: ∛x = -2 has the valid solution x = -8, since cube roots of negative numbers exist.
Question 14 of 20 · Multiple Choice
A driver covers 150 miles. If she drove 10 mph faster, the trip would take half an hour less. Solving 150/r - 150/(r + 10) = 1/2 gives r² + 10r - 3000 = 0. What is her speed?
Answer: A
r² + 10r - 3000 = (r + 60)(r - 50), so r = 50 or r = -60. Both satisfy the equation, so neither is extraneous, but a speed cannot be negative, so the context rules out -60. Check: 150/50 - 150/60 = 3 - 2.5 = 0.5. Choice B is the faster speed, r + 10, not her actual speed. Choice D reports the algebra without interpreting it.
Question 15 of 20 · Short Answer
Solve 2/(x + 3) + 1 = 5/(x + 3). State the excluded value and check your answer.
Excluded value: x = -3. Multiply every term by (x + 3): 2 + (x + 3) = 5, so x = 0. Check: 2/3 + 1 = 5/3 and 5/3 = 5/3. The candidate is not excluded and it checks, so it is a solution.
Question 16 of 20 · Short Answer
Solve √(x + 10) = x - 2. Label each candidate as a solution or extraneous.
Square: x + 10 = x² - 4x + 4, so x² - 5x - 6 = 0 and (x - 6)(x + 1) = 0. Candidates: 6 and -1. x = 6: √16 = 4 and 6 - 2 = 4, solution. x = -1: √9 = 3 but -1 - 2 = -3, extraneous. The solution is x = 6.
Question 17 of 20 · Short Answer
Solve x²/(x + 3) = 9/(x + 3).
Excluded value: x = -3. Multiply by (x + 3): x² = 9, so x = 3 or x = -3. Reject x = -3, because it makes the denominator 0. The solution is x = 3. Check: 9/6 = 9/6. The extraneous value came from multiplying both sides by x + 3, which equals 0 when x = -3.
Question 18 of 20 · Short Answer
Give an example of a rational equation that has one true solution and one extraneous solution, and explain how the extraneous solution arises.
Sample answer: x/(x - 3) = 3/(x - 3) + x. Excluded value: x = 3. Multiply every term by (x - 3): x = 3 + x(x - 3), so x² - 4x + 3 = 0 and x = 1 or x = 3. Check x = 1: 1/(-2) = -1/2 and 3/(-2) + 1 = -1/2, so it is a solution. x = 3 is extraneous, because it makes the denominators 0. It arises because multiplying both sides by x - 3 multiplies by 0 when x = 3, and the new equation is true there even though the original equation is undefined.
Question 19 of 20 · Short Answer
Solve √(5x - 1) = √(x + 7).
Square both sides: 5x - 1 = x + 7, so 4x = 8 and x = 2. Check: √9 = 3 and √9 = 3. Also confirm both radicands are not negative at x = 2 (9 and 9), so the solution is valid.
Question 20 of 20 · Short Answer
Sam can mow a large lawn alone in 4 hours. Working together, Sam and Ana mow it in 3 hours. Write and solve a rational equation to find how long Ana would take alone.
Let a = Ana's time alone in hours. Sam mows 1/4 of the lawn per hour and Ana mows 1/a. Together: 1/4 + 1/a = 1/3. Multiply by 12a: 3a + 12 = 4a, so a = 12 hours. Check: 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3. It makes sense that Ana is slower than Sam, since the pair saves only 1 hour compared with Sam alone.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What is an extraneous solution?
An extraneous solution is a value that comes out of a correct solving process but does not satisfy the original equation. It solves a later equation in the chain (such as the squared equation or the equation after clearing denominators) but not the one you started with. It is not a calculation mistake: it is produced by a step that can add solutions.
Which steps can create extraneous solutions?
Two steps in this standard: multiplying both sides by an expression that can equal 0 (as when clearing denominators in a rational equation), and squaring both sides (or raising them to any even power) in a radical equation. Adding, subtracting, and multiplying or dividing by a nonzero number never create extraneous solutions. Cubing both sides does not either.
Why doesn't cubing both sides create extraneous solutions?
Every real number has exactly one real cube root, so a³ = b³ happens only when a = b. Squaring is different: a² = b² is true when a = b and also when a = -b. That second case is where extraneous solutions come from.
What is the difference between an excluded value and an extraneous solution?
An excluded value is any x that makes a denominator of the original equation 0; you can list these before solving. An extraneous solution is a candidate produced by the solving process that fails the original equation. In a rational equation, extraneous solutions are usually excluded values that reappear after clearing denominators. In a radical equation, they are values where the two sides are opposites.
Do I have to check my answers if I made no mistakes?
Yes, whenever you squared both sides or multiplied by an expression containing the variable. Those steps can add solutions even when every calculation is correct, so the only way to find extraneous solutions is to substitute each candidate into the original equation. Checking in a later equation will not catch them, because every candidate satisfies the later equation by construction.
What are the common mistakes?
Squaring before isolating the radical, and writing (√x + 2)² as x + 4
Squaring a binomial side incorrectly, for example (x - 5)² as x² - 25
Multiplying only some terms by the common denominator
Checking candidates in the squared equation instead of the original
Listing both roots of the resulting quadratic as the final answer
Can a rational or radical equation have no solution?
Yes. If every candidate is extraneous, the equation has no solution. For example, x/(x - 5) - 3 = 5/(x - 5) produces only x = 5, which is excluded, and √(x - 3) + 5 = 2 requires a square root to equal -3, which is impossible. It is worth spotting the second kind before doing any algebra.
Is a negative answer in a word problem an extraneous solution?
Not necessarily. If the value satisfies the equation, it is a true solution of the equation. It may still be rejected because it does not make sense in the context, such as a negative speed or time. Encourage students to name which kind of rejection they are making: "fails the equation" (extraneous) or "fails the context."
Is HSA.REI.A.2 on the SAT?
Yes. The Advanced Math domain of the digital SAT includes nonlinear equations in one variable, and rational and radical equations appear in that category. Questions often give the solving work or the candidates and ask which value is a solution, so the checking habit is directly useful.
What comes after HSA.REI.A.2?
Students use the same ideas when they solve equations by graphing (HSA.REI.D.11), where an intersection of two graphs shows a true solution, and when they study rational and radical functions, domains, and inverse functions in Algebra II and Precalculus. The habit of asking "which step could add or lose solutions?" also matters when solving logarithmic and trigonometric equations later.
07
Related Standards
6 standards
These standards connect to HSA.REI.A.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.EE.A.2Prerequisite
Use square and cube root symbols to solve x² = p and x³ = p
