Reversible Steps: Losing, Creating, and Excluding Solutions
Mr. Okafor's algebra 2 class is solving equations, and he collects three exit slips. On item 1, 'Solve x² = 6x,' Leila writes x²/x = 6x/x, so x = 6, and boxes 6. On item 2, 'Solve x/(x−3) = 3/(x−3) + 2,' Marcus multiplies both sides by (x−3), obtains x = 3 + 2(x−3), so x = 3, and boxes 3. On item 3, 'Solve √(2x+5) = x − 1,' Tasha squares both sides, gets 2x+5 = x²−2x+1, so x²−4x−4 = 0, x = 2 ± 2√2, and boxes both values. All three students execute procedures cleanly when the structure is pointed out, but the class average on the unit test's equation items was 58%. Mr. Okafor has 25 minutes of reteach time before the retake and must choose one organizing idea rather than re-teaching three procedures separately. He also needs a quick way to tell whether the errors are procedural or conceptual.
1. For each slip, give the correct solution set and name the exact step that produced the error.
Read the worked answer
Leila (lost root). x² = 6x → x² − 6x = 0 → x(x − 6) = 0 → x = 0 or x = 6. Dividing by x presumes x ≠ 0; since x = 0 is a solution, the division silently discards it. Correct set: {0, 6}.
Marcus (excluded value). Multiplying by (x − 3) is legitimate only on the domain x ≠ 3. The algebra gives x = 3 + 2x − 6 → x = 3. But x = 3 makes both denominators zero, so the original equation is undefined there. Check: at x = 3 the equation reads 3/0 = 3/0 + 2, which is meaningless. Correct set: no solution (empty set).
Tasha (extraneous root). Squaring gives x² − 4x − 4 = 0, so x = 2 ± 2√2. The original requires x − 1 ≥ 0, i.e. x ≥ 1. Check x = 2 + 2√2 ≈ 4.83: LHS = √14.66 ≈ 3.83, RHS ≈ 3.83 ✓. Check x = 2 − 2√2 ≈ −0.83: RHS = −1.83, LHS ≥ 0, so it fails. Correct set: {2 + 2√2}.
The three errors are different symptoms of one cause: a step that is not reversible on the domain of the equation.
2. State the single organizing idea that unifies all three errors, and write the sentence you would put on the board.
Read the worked answer
Organizing idea: every step in solving an equation must be reversible - it must preserve the solution set. Dividing by an expression that might be zero is not reversible (it deletes solutions). Squaring is not reversible (it merges x − 1 with −(x − 1) and can invent solutions). Multiplying by an expression that might be zero is not reversible at the zeros of that expression (it can invent solutions where the original was undefined).
Board sentence: 'Before dividing, ask whether the divisor can be zero. Before squaring or clearing a denominator, record the restrictions and check every candidate in the original equation.'
A useful secondary framing: reversible steps need no checking; irreversible steps require a case split (divide by x only after separating the x = 0 case) or a verification pass (squaring, clearing denominators with variables). That framing converts three memorized warnings into one testable criterion students can apply to unfamiliar equations.
3. Describe a 25-minute reteach plan built on this idea, with one check-for-understanding item per error type and a way to distinguish conceptual from procedural errors.
Read the worked answer
Plan (25 min).
- 4 min: Post the three slips and ask pairs to find the line where 'information was lost or invented,' not the arithmetic slip.
- 8 min: For each equation, record domain restrictions first, solve, then verify candidates by substitution. Have students articulate the case split for x² = 6x.
- 8 min: Three new items, one per mechanism: (a) x³ = 4x (roots 0, 2, −2 - a division-by-x trap that also has three roots); (b) (x + 2)/(x − 5) = 7/(x − 5) → x + 2 = 7 → x = 5, excluded, no solution; (c) √(x + 3) = x − 3 → x² − 7x + 6 = 0 → x = 1 or 6, and x = 1 fails (RHS negative), so {6}.
- 5 min: Exit ticket with (a) and (c) only.
Procedural vs. conceptual. Procedural gaps disappear once the structure is named: the student can redo item (a) correctly after the prompt 'can x be zero?' Conceptual gaps persist across representations - the student still divides by x on a fresh item, cannot explain why x = 3 fails, or squares without any thought of checking. Diagnostic move: ask a student to explain the failure verbally and to build an equation whose only candidate is extraneous. An explanation that names reversibility signals concept; a recitation of 'always check your answers' signals a procedure memorized without meaning.