Additional Mathematics: Recheck Corrected Working

Correcting an Additional Mathematics mistake is not the same as finishing the correction. You can fix an algebraic sign, replace a wrong derivative, restore a missing solution or re-enter a calculator expression, yet leave the rest of the working attached to the old answer. This guide shows how to check corrected working without either trusting it too soon or solving the entire question again.

For Additional Mathematics revision, exam practice and worked solutions, the useful question is not simply “Does my answer now look right?” It is “What changed, which calculations used it, and what evidence shows that the corrected solution still satisfies the original question?” That question connects algebra, equations, trigonometry, calculus, coordinate geometry, exact answers and calculator checks into one practical examination habit.

Post-repair regression testing is the name used here for that habit: confirm the correction itself, recheck the parts it could affect, and check that previously valid work has not been damaged or silently left out. It is a teaching adaptation, not an official Additional Mathematics topic or a promise of any examination grade. The examples are original, and Alicia, Tricia and Kai Kai are fictional learners used to make different checking decisions visible.

This is a global guide. Match every practice task to your own qualification, examination year, calculator rules and required answer form. For example, the Cambridge IGCSE Additional Mathematics overview emphasises mathematical reasoning and structured arguments, but it does not make the assessment arrangements of every other Additional Mathematics course identical.

The 50-second route

Find the changed line. Say what the correction actually changed: a value, a formula, a domain, a branch, a diagram interpretation or a calculator setting. Check that correction against the original question. Then follow only the calculations and conclusions that used the old state. Do not preserve them just because they were checked before the edit.

Check more than the final number. A root must satisfy the original equation; a complete solution set also needs a reason that no roots were missed. A stationary point needs the correct function value and the classification requested. A corrected parameter may require a different case. A decimal comparison needs an appropriate precision rule.

Stop when the evidence is complete enough for the claim you are making. Reopen the check when another material edit occurs. In the final minutes, write one clear current value, verify its first important use and confirm the requested answer form. The aim is a trustworthy continuation, not an endless checking ritual.

Choose your reading route and open the chapter map

Repair just made: read Chapters 1–4, then choose the relevant mathematics example. Equations or missing roots: start with Chapters 5–11. Trigonometry or calculus: use Chapters 12–16. Calculator or numerical disagreement: read Chapter 18. Teaching and deliberate practice: begin with the integrated case and the diagnostic laboratory.

A corrected line creates a new checking question
Write a change record that remains useful
Find what the change can actually affect
Choose evidence that could catch the old mistake
Linear algebra and linked calculations
Rational equations and cancelled factors
Radicals, squaring and candidate validation
Logarithms and exponential substitutions
Modulus branches and inequality boundaries
Parameters that change the type of a question
Returning from a substituted variable
Trigonometric equations and complete angle sets
Multiple angles, division and lost solutions
Derivatives, stationary points and tangents
Integration, constants and signed area
Optimisation and feasible global answers
Coordinate geometry: equations must agree
Exactness, rounding and calculator handoffs
When correcting one error exposes another
Build a small but useful retest set
Use the method under examination time pressure
An integrated quadratic correction case
Ten diagnostic exercises with explained answers
A practical lesson and learner-specific routes
What one successful recheck does and does not show
Teaching guide, evidence notes and final operating routine

1. A corrected line creates a new checking question

Alicia has found the mistake. That matters: she is no longer guessing at a vague failure. In the middle of a quadratic solution, she copied a coefficient incorrectly. She replaces it, recomputes the next line and feels ready to continue. Three lines further down, however, the old coefficient remains inside a formula that she had already boxed. The page now contains two different versions of what is supposed to be one calculation.

There is nothing mysterious about this failure. Her first check asked whether the coefficient was correct. Her next job was different: establish whether the repaired solution was internally connected to that corrected coefficient. One accurate replacement cannot reach backwards through the page and change every other use automatically. On paper, the solver must do that work deliberately.

Tricia sees the same risk and adopts the opposite habit. Every time she changes anything, she solves the entire question again. That can catch mistakes, but it may also repeat safe work, consume time and create new opportunities to copy something incorrectly. Neither “I changed it, so it is fixed” nor “I changed it, so nothing is trustworthy” is a satisfactory default.

The central proposition of this guide is therefore narrow: a material correction should be followed by evidence proportionate to what changed and what depends on it. The proposition is not a licence to skip necessary proof. It is a way to organise proof, substitution, recomputation and consistency checks so that the most relevant questions are answered first.

The language has a useful origin in software testing. The ISTQB Foundation Level syllabus, section 2.2.3, distinguishes checking that a defect has been fixed from checking for adverse consequences of a change. We borrow that distinction, not a software certification process. Here, the objects being checked are equations, values, conditions and conclusions in a student’s worked solution.

Call the first job repair confirmation: did the change correct the specific fault? Call the second job post-repair checking: does the relevant surrounding solution still work after that change? Both questions may be answered by one well-chosen mathematical argument, but they are not logically identical. A correct derivative, for example, confirms one repaired formula. It does not by itself establish that the old stationary-point coordinates remain correct.

This also explains the relationship with Repair Sequencing. That guide asks which confirmed error to fix first. Here we begin after a correction has been made and ask what must be checked before later work can use it. When the next check reveals another fault, return to repair sequencing rather than pretending that the testing stage has somehow repaired it.

There are three possible outcomes of a recheck. A result can pass with sufficient evidence for its intended use. It can fail, in which case the exact failure should be recorded. Or it can remain unresolved because the chosen test is not strong enough. “Unresolved” is an honest mathematical status. It is better than changing a number merely because the student dislikes uncertainty.

Suppose a corrected expression agrees with the original expression at x = 1. That is one observation. It may catch a simple constant error, but it may fail to catch an incorrect coefficient multiplying x − 1. The test has not necessarily failed; it has answered a smaller question than the learner hoped. Good checking begins by stating that smaller question precisely.

Likewise, substituting a proposed solution into the original equation can establish that the proposed value works. It cannot show that there is no second solution somewhere else. Membership and completeness must not be confused. A solver who finds one valid trigonometric angle has not necessarily finished an interval question. A solver who verifies one stationary point has not necessarily found a global maximum.

Alicia’s new habit is not to create a large administrative system. After an important edit, she writes a short question beside it: “What used this?” Tricia adds a second question: “What did not use it?” Kai Kai adds a third: “Could this test still pass with the old mistake?” These questions create a practical balance between blind trust and excessive repetition.

Use the approach when the change has mathematical consequences. Correcting a spelling error in the word “therefore” does not require a new quadratic solution. Correcting a minus sign in the quadratic does. Changing a final approximation symbol may be a presentation repair; changing the number that was rounded may require numerical recomputation. The work should follow the change’s meaning, not the amount of ink altered.

2. Write a change record that remains useful

A useful change record tells a returning reader what the solution now means. It is not a diary of every thought. It should fit into ordinary mathematical working and survive a brief interruption. For a substantial correction, record the corrected object, the reason for the change, the conditions attached to it, the first important calculation that uses it, and the remaining check.

For example: “k = −2 from the original condition; replace k = 2 in the tangent calculation; recheck the point and line.” That note is more useful than “fixed sign”. It identifies the object, distinguishes the new value from the old one, gives the source of authority and tells the learner where to look next. It also prevents the old boxed result from silently remaining in use.

The corrected object need not be a number. It may be a domain such as x > 1, an interval such as 0 ≤ θ < 2π, a definition such as u = 2^x, or a statement such as “the lower endpoint is excluded”. Conditions often govern more working than a single arithmetic value. They deserve a visible record when they change.

Consider a logarithmic equation. A student corrects the expanded polynomial but leaves the domain written as “all real x”. The polynomial may now be accurate while the candidate filtering remains wrong. A change record that only says “algebra fixed” misses this second responsibility. A better record identifies both the regenerated candidates and the original logarithmic restrictions under which candidates may be accepted.

There is a useful distinction between the problem’s conditions and the student’s record of those conditions. The examination question has not changed merely because the learner copied an interval incorrectly. The repair restores the learner’s record to the original condition. This distinction matters because otherwise “the domain changed” can sound like permission to alter the problem. It means the written domain was corrected, not that the student chose a more convenient domain.

When the source is a diagram, record the exact relation rather than an impression. “AB is perpendicular to CD” is checkable. “The angle looks about right” is not equivalent evidence. If a diagram is explicitly not to scale, the visual sketch cannot override stated lengths, coordinates or angle relations. A corrected diagram label should trigger checks on equations that used that label, not on unrelated parts merely because they share the page.

Tricia uses a compact four-part note in practice: “Changed / Used by / Still valid / Recheck”. She does not reproduce this table during every timed solution. Training with the categories teaches her to see them quickly. For a corrected derivative, she can write: “Changed: f′. Used by: stationary x and tangent slope. Still valid: original f. Recheck: coordinates and classification.”

The “still valid” entry is especially valuable. It prevents indiscriminate restarting. If the original function was copied correctly, it remains a trustworthy source for recalculating coordinates. If the correction only changes a final decimal, the exact expression above it may remain valid. But preservation needs a reason. A line is not protected merely because it is earlier, neater or already ticked.

Keep old and new values distinguishable. A light single strike-through and a clear replacement may be enough. Writing a minus sign over a plus sign until neither is readable creates another problem. When the edit is hard to read, rewrite the short decisive line rather than polishing the whole page. The aim is one clear current state, not a pristine notebook.

Calculator results also need a source. If the displayed number belongs to an earlier expression, changing the handwritten expression does not update the calculator. The change record should say which expression now produces the numerical result. A fresh entry from the corrected written expression is often more transparent than trying to remember which stored value or previous-answer register was used.

Do not over-record disposable mental arithmetic. Evaluating 2 × 3 + 1 as 7 can reasonably happen in one line. The record becomes important when a value will be reused, a restriction controls a branch, or a correction affects several later parts. The more a result is reused, the more useful it is to identify where it came from and what still needs checking.

There is no need to invent a certainty score. “Checked in the original equation” communicates actual evidence. “Ninety-five per cent certain” usually does not, unless there is a defensible probabilistic model behind it. The student should record what was tested and what the test establishes. That makes later decisions more reliable than collecting vague confidence labels.

Before leaving a partly repaired question, write the smallest useful restart note: the current equation, the active branch and the next unresolved check. On return, rebuild the mental picture from that record. This complements Working-State Synchronisation: the page and the student’s next action should describe the same mathematics.

3. Find what the change can actually affect

A dependency is a mathematical use, not a position on the page. If a value of k enters both a gradient formula and an area formula, both calculations depend on k even if they are several pages apart. If an independent distance calculation appears immediately underneath, it may not depend on k at all. The recheck should follow those relationships.

Start with the corrected result and identify its direct users. Then ask what uses each of those users. In practice, you rarely need to draw a complete graph. A short chain such as “parameter → equation → roots → coordinates → line” is enough to reveal that an unchanged-looking final line may need retesting. The corrected parameter has not automatically repaired that line’s provenance.

There are four useful actions. Recompute a result when an input it actually consumed has changed. Revalidate a result when its algebra may survive but its conditions or interpretation need checking again. Preserve a result when a clear argument shows that the correction does not affect it. Retire a result when it belongs to an abandoned branch or no longer answers any live part of the problem.

For example, correcting the coefficients of a quadratic normally requires regenerating its roots. Correcting only a domain restriction may leave the algebraic candidates unchanged while requiring the candidate list to be filtered again. These are different jobs. A solver who regenerates everything after every domain edit wastes work; a solver who retains the old accepted list without refiltering risks missing the actual consequence.

Independence also has to be established at the right level. Two calculations can look different but share the same corrupted input. An algebraic solution and a calculator solution are not independent confirmations if both use the same miscopied equation. Before using agreement as evidence, trace them back to the original problem rather than to a convenient intermediate line.

Conversely, some dependents remain numerically unchanged for a genuine mathematical reason. Suppose a curve is shifted vertically by a constant. Its derivative does not change. A correction to the constant term therefore changes function values and tangent intercepts but not the derivative formula or the x-coordinates of stationary points. Preservation here rests on differentiation of a constant, not on the fact that the old answer happens to look familiar.

This example gives a valuable principle: a dependency warning is not the same as a demand to recalculate blindly. It asks whether the change can alter the particular claim. If a proved invariant protects the claim, document that briefly and keep it. If no such argument is available, recompute or revalidate. “Probably unaffected” is not a substitute for identifying the invariant.

Alicia practises with a quadratic written as f(x) = x² − 4x + b. Correcting b from 5 to 3 changes f(2) from 1 to −1. The derivative remains 2x − 4, so the stationary x-coordinate remains 2. The stationary point, however, is not unchanged: its y-coordinate has changed. One label, “stationary point”, contains two claims with different dependencies.

The same issue appears in geometry. A line may keep its gradient after a coordinate correction while its intercept changes. A circle’s centre may remain fixed while its radius changes. A trigonometric identity may remain correct while the interval over which its solutions are required changes. Retesting should separate these claims instead of treating an entire paragraph as either wholly valid or wholly invalid.

When a correction changes mathematical type, widen the scope. A leading coefficient becoming zero can turn a quadratic into a linear equation. A corrected denominator can make a formerly allowed value undefined. A parameter threshold can change which branch applies. In those situations, retaining the old method and merely inserting the new number may be invalid.

Pay attention to joins, where two calculations combine. Suppose a final area uses a width from one branch and a height from another. If only the width changes, the height may remain valid, but the area must still be recomputed. The join is an important recheck point because it can combine one current result with one stale result without making the mismatch visually obvious.

Do not use a passed midpoint as an automatic proof that all earlier work is correct. Two earlier mistakes can cancel at that midpoint, or the midpoint test can be insensitive to the fault. A successful test establishes the property it actually checks. It may narrow a search when the relevant assumptions are justified, but it is not a universal binary-search certificate for a handwritten proof.

The relationship with Dependency Chains is practical: use the chain to select the scope of the recheck. The purpose here is not to label every arrow. It is to decide which old conclusions can still be carried forward and which need fresh evidence after the edit.

4. Choose evidence that could catch the old mistake

A test is useful only in relation to a claim. “I checked it” is incomplete until the solver can explain what was checked and what failure the check could reveal. Reading the same line three times may be appropriate for a transcription error, but it does not automatically test a missing branch, a forbidden value or a wrong interpretation of the question.

Begin by naming the old fault. Suppose the student incorrectly expanded (x + 2)² as x² + 4. The corrected form is x² + 4x + 4. Testing only x = 0 cannot distinguish the wrong expansion from the correct one because both give 4. The chosen value lies exactly where the missing term vanishes. At x = 1, the original expression gives 9 while the wrong expansion gives 5, so that test can expose this particular fault.

However, the passing x = 1 test still does not prove a general identity. The proof is distributive expansion: (x + 2)(x + 2) = x² + 2x + 2x + 4 = x² + 4x + 4. The numerical test is a quick diagnostic and the expansion establishes the identity for every real x. Different kinds of evidence do different jobs.

There is a legitimate finite-testing result in a more restricted setting. If two polynomials of degree at most two agree exactly at three distinct inputs, their difference is a polynomial of degree at most two with three distinct roots, so the difference must be zero. But this reasoning requires the degree bound, exact agreement and distinct inputs. It does not license three arbitrary calculator samples for an unknown function or a rational expression with missing domain restrictions.

This distinction is worth teaching because it prevents both extremes. Numerical testing is not worthless; it can be excellent at revealing a specific error. Nor is it a universal replacement for algebraic proof. The learner should know when a theorem makes finite testing sufficient and when the samples only provide limited evidence.

Now consider the claim that x = 4 solves √(3x + 4) = x. Substitution gives √16 = 4, so membership is confirmed. To establish completeness, the student must also show how the candidate set was generated and filtered. The original equation implies x ≥ 0. Squaring gives x² − 3x − 4 = 0, whose candidates are 4 and −1. The negative candidate is rejected. Membership and completeness together establish the solution set.

A test can also be insensitive because the final output erases information. If a wrong value is −3 instead of 3, checking only its square gives 9 in both cases. If the question needs the original signed value, that test cannot settle the claim. A sign-sensitive original equation, interval condition or geometric orientation is needed.

Similar limitations apply to summaries. Two wrong roots can have the correct sum. A correct sum does not prove that both roots are individually valid. Checking the product adds another relation, and substituting each candidate into the original polynomial checks membership directly. The appropriate combination depends on the task and the transformations used.

Tricia learns to choose a small set of different checks rather than repeat one check many times. For a quadratic repair, she might confirm the coefficient from the question, verify the factorisation by expansion, substitute the resulting roots and check a stated domain. Each check targets a different possible failure. None is included merely to make the checklist longer.

When using approximations, define the comparison honestly. Exact symbolic equality and agreement to three significant figures are different claims. A calculator display that rounds two nearby numbers to the same value cannot establish exact equality. Conversely, a tiny decimal discrepancy may be irrelevant when the requested answer is rounded and the underlying exact calculation is sound.

Keep expected results independent of the disputed line where practical. If both the calculation and its check start from the same miscopied function, agreement says little about fidelity to the question. Return to the original statement, a separately established relation or an independently derived invariant. The Parallel-Route Isolation guide develops the broader issue of keeping methods from borrowing unverified states.

A useful final question is: “Would this check have failed before I corrected the mistake?” If not, the check may still test something valuable, but it is not evidence that this particular defect was removed. Add a discriminating check rather than counting the pass twice. That one habit turns vague reassurance into a testable account of what the repaired solution can now support.

5. Linear algebra and linked calculations

Start with a small example because the checking problem is easier to see when the algebra is familiar. Solve 3(2x − 1) − 2(x + 4) = 9, then use the result to calculate y = x² − 2x. Alicia expands the second bracket incorrectly as −2x + 8. She notices the sign error and changes +8 to −8. That is the repair. What must happen next?

The corrected expansion is 6x − 3 − 2x − 8 = 9. Combining like terms gives 4x − 11 = 9, hence 4x = 20 and x = 5. The original equation provides a direct check: 3(10 − 1) − 2(5 + 4) = 27 − 18 = 9. This check uses the original brackets, not the corrected expanded line, so it can expose a repeated expansion mistake.

The linked result is y = 25 − 10 = 15. If an old y-value remains on the page, it must be replaced or explicitly revalidated. A correct x-value is not permission to retain a y-value that was calculated from another x. The first important downstream use is therefore part of the recheck, not an optional extra.

There is also a completeness argument. After valid expansion and collection, the equation is 4x = 20. Its coefficient is nonzero, so it has exactly one real solution. Substitution establishes that 5 works; the equivalent linear equation establishes that there is no other real solution. In this simple case, the two responsibilities fit into very little working.

Now imagine that Tricia has independently calculated z = 7² − 3² elsewhere on the page. That result, z = 40, does not use x. It should not be recomputed just because it appears below the repaired equation. She can preserve it while rechecking y. This small example trains the distinction between chronological order and mathematical dependence.

The test record can be written in one sentence: “Expansion corrected; x = 5 passes the original equation; y recalculated as 15; independent z unchanged.” During an examination, even that sentence may be unnecessary if the mathematics already displays the evidence. In a practice review, however, it reveals whether the learner understood the scope of the correction.

Consider a second example involving simultaneous equations: 2x + y = 7 and x − y = 2. Adding the equations gives 3x = 9, so x = 3 and y = 1. Suppose Kai Kai originally copied the second equation as x + y = 2, then repairs the sign. The first equation remains correct. The old elimination result does not. Both variables need to be regenerated from the corrected pair.

Checking only 2x + y = 7 is inadequate because many pairs satisfy that one equation. Both original equations must hold: 2(3) + 1 = 7 and 3 − 1 = 2. The corrected pair must solve the system, not merely the equation that happened to be easiest to check. This is an early example of testing a joint claim.

A useful teaching question is: “Can you construct a wrong pair that passes your chosen check?” For the first equation, the pair (2, 3) passes because 4 + 3 = 7, but it fails the second because 2 − 3 = −1. The student can now see exactly why a single-equation test is too weak. The limitation is demonstrated rather than asserted.

There is no need to solve the entire system a second way after these checks if the original derivation is transparent and complete. The aim is not to maximise the number of methods used. It is to confirm that the repaired sign has been applied to the live system and that the resulting pair meets every required equation.

When the same value is used several times, follow the first use and every distinct important use. A corrected x may feed a square, a reciprocal and a logarithm. These are not interchangeable downstream checks: the reciprocal has a nonzero condition and the logarithm has a positive-argument condition. A simple source repair can therefore lead to different kinds of validation further down the solution.

This is where the small example grows into advanced examination work. The habit is unchanged: confirm the edit at its source, regenerate actual dependents, test the complete original requirement and preserve genuinely independent work. Later chapters add domains, cases, intervals and interpretation, but none removes this basic responsibility.

6. Rational equations and cancelled factors

Rational expressions make post-repair checking particularly important because simplification can hide an excluded value. The expression (x² − 9)/(x − 3) simplifies to x + 3 only when x ≠ 3. The simplified formula is easier to use, but it does not acquire the right to define the original expression at x = 3.

Solve (x² − 9)/(x − 3) = 8. The original restriction is x ≠ 3. Factorising the numerator gives (x − 3)(x + 3)/(x − 3) = 8. On the allowed domain, cancellation yields x + 3 = 8, so x = 5. Substituting into the original equation gives (25 − 9)/(5 − 3) = 16/2 = 8. The candidate is valid.

Suppose the learner first wrote the wrong factorisation, repaired it and obtained x = 5. What remains to be checked? The factorisation must agree with the original numerator; the cancellation must retain x ≠ 3; the candidate must satisfy the original equation; and the route must cover the allowed domain. The clean reduced equation alone does not display all four responsibilities.

A nearby variant makes the hidden restriction visible. Solve the same rational expression equal to 6. The reduced equation gives x + 3 = 6, hence x = 3. But x = 3 is excluded by the original denominator. Therefore the original equation has no solution. Substituting into the reduced expression is not an adequate final test because it tests an extension of the expression beyond its original domain.

This pair of examples is useful during training. The algebraic operation is the same in both cases, yet the acceptance outcome differs. A routine that automatically accepts the reduced equation’s answer will succeed on the first and fail on the second. The second item reveals whether the learner actually carried the domain through the repair.

Now solve 1/(x − 1) + 1/(x + 1) = 3/4. The original restrictions are x ≠ 1 and x ≠ −1. Combining the left side gives 2x/(x² − 1) = 3/4. Multiplying by 4(x² − 1), which is nonzero on the permitted domain, gives 8x = 3x² − 3. Thus 3x² − 8x − 3 = 0.

The factorisation is (3x + 1)(x − 3) = 0, so the candidates are x = −1/3 and x = 3. Both avoid the excluded values. The original-equation checks are short: at x = 3, 1/2 + 1/4 = 3/4; at x = −1/3, −3/4 + 3/2 = 3/4. Both candidates are accepted.

Imagine that Tricia corrects 3x² − 8x + 3 to 3x² − 8x − 3 but keeps one root from the old quadratic. Retesting the repaired polynomial catches the stale candidate. Checking the original rational equation adds protection against domain or denominator mistakes. These checks have different targets, even though both involve substitution.

Completeness follows from the reversible operations on the restricted domain and the complete factorisation of the resulting quadratic. There is no need to sample many additional real numbers. The algebra has already narrowed every possible original solution to the two candidates, and both have been tested. This is a proof of the solution set, not merely a collection of successful numerical tests.

Alicia’s practical record for rational equations has three parts: excluded values, transformed equation and accepted candidates. If a repair changes the denominator or a cancelled factor, she rechecks all three. If it changes only the final transcription of an already verified root, the scope may be smaller. The kind of edit determines which evidence needs refreshing.

Do not confuse ordinary algebraic cancellation with the hidden-error cancellation discussed elsewhere in the series. Cancelling a common nonzero factor is a valid operation under its condition. Two wrong steps accidentally offsetting one another is a different issue. Precise language helps students avoid becoming suspicious of legitimate simplification while missing actual domain loss.

The final question before continuing is: “Does the current simplified expression represent the original problem on the domain I am using?” If the answer is conditional, keep the condition visible. This is a direct application of Constraint Integrity, now used specifically to decide whether a correction has been safely incorporated.

7. Radicals, squaring and candidate validation

Squaring an equation is a useful operation, but it can lose sign information. From A = B we may infer A² = B²; the reverse implication requires additional conditions. In real-number work, A² = B² allows A = B or A = −B. A repaired squared equation can therefore be correct while its candidate list still contains values that do not solve the original equation.

The OpenStax treatment of radical equations explicitly addresses extraneous candidates and checking in the original equation. The important point for this guide is that an extraneous candidate need not indicate careless algebra. It can arise because the chosen transformation gives a necessary condition without preserving full equivalence.

Take the original example √(3x + 4) = x. The square root must be defined, so 3x + 4 ≥ 0. Since the principal square root is nonnegative, any solution also satisfies x ≥ 0. The latter condition already implies the former here. Squaring gives 3x + 4 = x², hence x² − 3x − 4 = 0.

Factorisation gives (x − 4)(x + 1) = 0. The candidates are 4 and −1. At x = 4, the original equation gives √16 = 4, which is true. At x = −1, it gives √1 = −1, which is false. Thus the solution set is {4}. Both candidates solve the squared equation; only one solves the original equation.

Suppose Kai Kai repairs the factorisation and then ticks both candidates because each makes x² − 3x − 4 equal to zero. His check confirms the factorisation and root extraction. It does not confirm the original radical equation. The test is useful, but he has asked it to establish a claim it cannot establish.

The correction record should distinguish “candidate generation repaired” from “candidate validation complete”. After a material repair to the squared equation, regenerate the candidates and test every remaining candidate in the original relation. Do not carry an earlier acceptance decision into a new candidate list without checking whether the original sign condition still holds.

Now consider √(x + 5) = x − 1. Any real solution requires x ≥ 1. Squaring gives x + 5 = x² − 2x + 1, hence x² − 3x − 4 = 0. The same quadratic appears, with candidates 4 and −1. At x = 4, √9 = 3 and x − 1 = 3, so it works. The candidate −1 violates x ≥ 1 and fails the original equation.

The two radical equations produce the same squared polynomial, but they have different original forms and sign requirements. This is a good reason not to use a remembered polynomial as the sole source of truth. A student returning after an interruption should read the original equation before validating the candidates. Similar transformed work can belong to different original problems.

A harder recheck concerns √(x + 6) = 2 − x. The right side must be nonnegative, so x ≤ 2, and the radicand requires x ≥ −6. Squaring gives x + 6 = x² − 4x + 4, hence x² − 5x − 2 = 0. The candidates are (5 ± √33)/2. Only the smaller candidate lies in the required interval [−6, 2].

To verify the accepted candidate without relying on a decimal, let r = (5 − √33)/2. It satisfies the squared equation. It also satisfies r ≤ 2 and r ≥ −6, so both √(r + 6) and 2 − r are nonnegative. Their squares are equal; therefore the original nonnegative quantities are equal. This is a valid sign-aware argument. It is stronger than merely observing that two rounded calculator displays look similar.

The rejected candidate (5 + √33)/2 is greater than 2. Its right side, 2 − x, is negative, while a principal square root cannot be negative. A direct sign argument rejects it without a long numerical substitution. Good rechecking is not synonymous with expensive arithmetic; sometimes the strongest evidence is a simple structural condition.

After the repair, completeness is established because every original solution must appear among the roots of the squared quadratic, and the sign conditions have identified exactly which candidates return to the original equation. If additional squaring steps were used, repeat this reasoning for the original relation rather than assuming the last transformed equation preserved all earlier restrictions.

This chapter also sets a boundary for the phrase “check every answer”. It means every candidate that could be presented as a solution, not every real number in the domain. The derivation supplies the candidate universe; validation filters it. A good post-repair routine keeps those two parts connected.

8. Logarithms and exponential substitutions

Logarithmic work often contains a subtle mismatch between an equation’s algebraic form and its original domain. A correction to the algebra does not automatically repair that mismatch. Start by writing the positive-argument conditions for the original real logarithms, then carry them through simplification.

Solve log₂(x − 1) + log₂(x + 1) = 3. Both arguments must be positive, so x − 1 > 0 and x + 1 > 0. Together these give x > 1. Combining the logarithms on that domain gives log₂[(x − 1)(x + 1)] = 3. Therefore x² − 1 = 8 and x² = 9.

The algebraic candidates are x = 3 and x = −3. Only 3 satisfies x > 1. Checking it in the original equation gives log₂2 + log₂4 = 1 + 2 = 3. The candidate −3 does not produce two real logarithms in the original expression and must be rejected.

Suppose Alicia first writes x² + 1 instead of x² − 1, corrects the product and then accepts both ±3. The product repair is valid. The solution is still incomplete as a checked object because the acceptance stage ignored the original domain. The recheck must include both the corrected product and the original positive-argument restrictions.

Notice why checking only the combined product can mislead. At x = −3, the product (x − 1)(x + 1) is positive, so log₂(x² − 1) is defined. Yet the separate original arguments are both negative. The combined expression can have a broader real domain if the original restrictions are forgotten. A valid logarithm law was applied under conditions; those conditions must remain attached.

This is not a minor notation issue. It changes the set of allowed inputs. A post-repair check should therefore begin from the original logarithmic statement, not from the last simplified form. When a student says “but the calculator gives a value”, ask which expression was entered and whether it is the original expression on the same domain.

Exponential substitutions create a different return problem. Solve 2^2x − 5·2^x + 4 = 0. Let u = 2^x, so u > 0. The equation becomes u² − 5u + 4 = 0, or (u − 1)(u − 4) = 0. Thus u = 1 or u = 4.

Returning to x gives 2^x = 1 or 2^x = 4, hence x = 0 or x = 2. Checking the original equation gives 1 − 5 + 4 = 0 at x = 0 and 16 − 20 + 4 = 0 at x = 2. Because the exponential function with base 2 is one-to-one, each positive u-candidate has exactly one real preimage.

Tricia’s first draft ends with “x = 1 or 4”. She has solved correctly in u and then labelled the u-values as x-values. The repair is not just changing a letter at the end. She must execute the return map and recheck the original equation at the resulting x-values. A transformed-variable answer is not automatically an answer to the original question.

A variation tests the restriction u > 0. Solve 3^2x + 2·3^x − 3 = 0. With u = 3^x > 0, the quadratic is u² + 2u − 3 = (u + 3)(u − 1) = 0. The candidate u = −3 is impossible for a real exponential of base 3. The surviving candidate u = 1 gives x = 0.

Checking only the quadratic would accept both u-values. Checking only the label “positive exponential” without solving would not establish the final x-value. The complete route needs the definition of u, the quadratic candidates, the positivity filter and the return to x. A repair at any one of these interfaces should trigger a recheck of the later interfaces it can affect.

Do not invent a real logarithm of a negative u-candidate to force the return step. The substitution’s range has already ruled it out. Equally, do not introduce complex-number work unless the course and question actually require it. This guide’s examples are real-number problems, and their domain statements are part of the solution.

The final check can be compact: original domain, current transformed equation, permitted transformed candidates, original-variable answers and original-equation substitution. Each item protects a distinct handoff. This is how the learner avoids a correct quadratic calculation becoming an incorrect logarithmic or exponential answer.

9. Modulus branches and inequality boundaries

Absolute value questions make local conditions visible. The identity |2x − 3| = 2x − 3 holds when x ≥ 3/2; the identity |2x − 3| = 3 − 2x holds when x < 3/2. A correction inside one branch must be tested under that branch’s condition before its candidate enters the final answer.

Solve |2x − 3| = x + 1. The right side must be nonnegative, so every solution has x ≥ −1. In the branch x ≥ 3/2, the equation becomes 2x − 3 = x + 1, giving x = 4. This candidate satisfies the branch condition and the original equation: |8 − 3| = 5 = 4 + 1.

In the branch x < 3/2, the equation becomes 3 − 2x = x + 1. Thus 2 = 3x and x = 2/3. It satisfies the branch condition and the original equation because |4/3 − 3| = 5/3 = 2/3 + 1. The solution set is {2/3, 4}.

Suppose Kai Kai initially writes −2x − 3 in the negative branch, then corrects it to 3 − 2x. Retesting only the first branch contributes nothing to confirming this repair. He needs to regenerate the second branch’s candidate and validate that candidate against both its local condition and the original equation. The first branch can be preserved because its derivation did not use the corrected negative-branch expression.

Completeness is a separate check on the branch system. The cases x ≥ 3/2 and x < 3/2 cover every real input and do not overlap. The threshold belongs to the first branch. At x = 3/2, the original equation gives 0 = 5/2, which is false. Both branches have been solved as linear equations, so there are no additional solutions hidden at the boundary or in an unexamined interval.

Notice that x ≥ −1 was a necessary global restriction but not a complete solution. A learner who checks only that both candidates exceed −1 has not verified the equation. A learner who verifies both candidates but forgets an entire branch has not established completeness. The recheck should distinguish these different claims.

Now consider the rational inequality (x − 1)/(x + 2) ≥ 0. The denominator excludes x = −2. The numerator vanishes at x = 1. These two critical values divide the allowed number line into three open intervals. For x < −2, numerator and denominator are both negative, so the quotient is positive. For −2 < x < 1, the numerator is negative and the denominator positive, so the quotient is negative. For x > 1, both are positive.

The endpoint x = 1 is included because the quotient is defined and equals zero. The value x = −2 is excluded because the quotient is undefined. Therefore the solution is x < −2 or x ≥ 1, equivalently (−∞, −2) ∪ [1, ∞). These endpoint decisions are not decorations added after the main calculation; they are part of the inequality’s meaning.

Tricia corrects a sign-chart entry but leaves the final answer as [−2, 1]. A post-repair check should return to interval signs and endpoint status. Test values such as −3, 0 and 2 are useful confirmations: the respective quotients are 4, −1/2 and 1/4. Yet the reason one test value can represent each interval is that the linear factors do not change sign within it. It is not a general licence to test one arbitrary point in any interval of any function.

A boundary-only check can also fail. Excluding −2 and including 1 does not decide which open intervals belong in the answer. Conversely, a correct open-interval sign pattern can still produce an incorrect answer if an endpoint is attached wrongly. Retest the interval decisions and the boundary decisions as separate items.

The full Case Coverage guide develops branch completeness. In this post-repair setting, the specific question is whether the edited branch or threshold has changed the final union. One corrected inequality symbol can require a new boundary decision even when every interior calculation remains valid.

For teaching, place two scripts side by side: one with correct candidates but missing branch conditions, and one with correct conditions but an incomplete candidate list. Ask the learner which exact evidence is absent in each. This helps replace the vague instruction “check the cases” with visible, testable responsibilities.

10. Parameters that change the type of a question

A parameter correction can alter more than a numerical answer. It can change the kind of equation being solved. When a leading coefficient becomes zero, a quadratic formula that previously looked appropriate may no longer apply. Post-repair testing must therefore begin by asking whether the old method remains valid under the corrected parameter.

Consider (k − 1)x² − 2kx + (k + 1) = 0. A useful factorisation is (x − 1)[(k − 1)x − (k + 1)] = 0. Expanding confirms the coefficients: the middle term is −(k + 1)x − (k − 1)x = −2kx, and the constant term is k + 1. This factorisation is valid for every real k.

When k ≠ 1, the equation is quadratic and has candidates x = 1 and x = (k + 1)/(k − 1). These values are distinct because equality would require k + 1 = k − 1, which is impossible. The discriminant is 4k² − 4(k − 1)(k + 1) = 4, consistent with two distinct real roots when the leading coefficient is nonzero.

When k = 1, the original equation is −2x + 2 = 0, so its only solution is x = 1. The expression (k + 1)/(k − 1) is undefined at k = 1 and cannot be treated as a second root. The discriminant calculation by itself does not rescue the quadratic interpretation once the quadratic coefficient vanishes.

Suppose Alicia repairs a parameter value from k = 2 to k = 1. Under k = 2, the equation is x² − 4x + 3 = 0 with roots 1 and 3. Under the corrected k = 1, it is linear with only root 1. Retesting the old root 3 in the corrected original equation gives −6 + 2 = −4, so it fails. More importantly, the old two-root architecture must be retired.

This is a strong example of why “replace the number everywhere” can be inadequate. The correction changes which operations are legal and which outputs should exist. A result that disappears because a branch no longer applies does not need a local numerical patch. It needs to be removed from the current solution set.

Now imagine the parameter is corrected from k = 2 to k = 0. The equation becomes −x² + 1 = 0, giving x = ±1. The general formula produces 1 and (0 + 1)/(0 − 1) = −1. The equation remains quadratic, but the old root 3 is still invalid. Here the method survives while one root changes. Comparing this case with k = 1 teaches the difference between changed data and changed mathematical type.

The post-repair test set should include the parameter’s defining condition, the actual substituted equation, the nonzero condition of any formula used, the current candidate set and the original-equation checks. When k = 1, a direct linear solution supplies completeness. When k ≠ 1, the factorisation supplies completeness and the nonzero leading coefficient supports the two-root claim.

Do not treat the special case as a nuisance to be squeezed into a formula. It is a legitimate part of the parameter space. The Exceptional-Case Control guide explains the broader issue; here the correction itself is the trigger to reopen that issue.

A second parameter example illustrates a changed domain rather than changed degree. Suppose y = 1/(x − k) and a downstream evaluation uses x = 2. If k is corrected from 1 to 2, the old value y = 1 is not merely inaccurate; the requested expression is now undefined. Re-entering it into a calculator cannot supply a legitimate real value. The domain check must happen before numerical reuse.

Parameter testing also needs restraint. A few examples such as k = 0, 1 and 2 are excellent for exposing a forgotten special case, but they do not prove a formula for every real k. The symbolic factorisation and explicit case distinction provide that proof in the quadratic example. Samples support diagnosis; the general argument carries the universal claim.

Tricia’s useful habit is to write the parameter restriction beside any formula with that parameter in a denominator. After an edit, she rereads that condition before substituting. Kai Kai’s useful habit is to substitute the corrected parameter into the original equation once. That direct view often reveals a change of type that a long general formula conceals.

11. Returning from a substituted variable

A substitution can simplify a difficult equation while creating a new responsibility: the solution must return to the original variable. Post-repair checking should treat that return as a real mathematical stage. Correct work in u is not automatically correct work in x, and a corrected u-value may have more than one, one or no real x-values associated with it.

Solve x⁴ − 5x² + 4 = 0 over the real numbers. Let u = x², so u ≥ 0. The transformed equation is u² − 5u + 4 = 0, which factors as (u − 1)(u − 4) = 0. Thus u = 1 or u = 4. Both satisfy the substitution’s range restriction.

Returning to x gives x² = 1 or x² = 4. Therefore x = −1, 1, −2 or 2. The final set is {−2, −1, 1, 2}. Substitution in the original quartic confirms that ±1 give 1 − 5 + 4 = 0 and ±2 give 16 − 20 + 4 = 0.

Suppose Kai Kai repairs a wrong transformed coefficient and now has u = 1 and 4, but retains only x = 1 and 2. The transformed equation is fixed. The return map is incomplete because x² is not one-to-one on all real numbers. A test that only squares 1 and 2 passes while still missing −1 and −2.

The completeness check asks how many original-variable values correspond to each accepted transformed value. For a positive u in u = x², there are two real values, ±√u. For u = 0, there is one real value. For u < 0, there is none. These cases should be understood before the final answer set is assembled.

A variation makes the impossible case visible. Solve x⁴ + x² − 6 = 0. With u = x² ≥ 0, the equation becomes (u + 3)(u − 2) = 0. The candidate u = −3 is rejected because no real square is negative. The accepted candidate u = 2 gives x = ±√2. Both satisfy the original equation because 4 + 2 − 6 = 0.

A student who automatically takes square roots of every transformed candidate may invent inappropriate complex answers or silently discard a sign. The relevant real-number restriction should be applied at the transformed stage, followed by the complete real return map. The repair is finished only when the final values belong to the original problem.

Now compare this with the exponential substitution u = 2^x from the previous chapter. A positive u has one real preimage because the base-2 exponential function is one-to-one. The two substitutions have different return behaviour. A generic instruction such as “always write plus or minus when returning” is therefore wrong. The correct rule comes from the mapping, not a memorised habit.

A restricted original domain can change the return count too. If x⁴ − 5x² + 4 = 0 is asked only for x > 0, the final answer is {1, 2}. The negative values remain algebraic roots but are not admissible for that question. Post-repair testing must use the domain actually stated, not the domain of a familiar example.

Alicia labels the current variable beside candidate lists. “u-candidates: 1, 4” cannot be mistaken for “x-solutions: 1, 4” when she returns after an interruption. This small notation choice protects the handoff. It is particularly useful when the transformed equation has the same visual shape as a familiar equation in x.

For more elaborate substitutions, record the inverse relation or return procedure before solving. If t = x − 2, returning is x = t + 2. If u = 1/x, the original restriction x ≠ 0 matters and the transformed value u = 0 cannot correspond to a finite real x. The point is not to collect many substitution tricks, but to know what each change of variable preserves and what it excludes.

The recheck can be stated as four questions: is the transformed equation correct, are its candidates permitted by the mapping, has every valid original-variable value been recovered, and do those values meet the original equation and domain? A repair at one stage does not erase the obligations at the others.

12. Trigonometric equations and complete angle sets

Trigonometric checking must distinguish a correct principal value from a complete answer over an interval. A repaired identity or quadratic in sin θ can still lead to an incomplete angle list. The interval, angle unit, function range, periodicity and endpoint status all belong to the current solution.

Solve 2sin²θ − 3sin θ + 1 = 0 for 0 ≤ θ ≤ 2π. The interval is in radians. Factorisation gives (2sin θ − 1)(sin θ − 1) = 0, so sin θ = 1/2 or sin θ = 1. Both values lie within the range [−1, 1] and therefore can produce real angles.

Within one full turn, sin θ = 1/2 gives θ = π/6 and 5π/6. The equation sin θ = 1 gives θ = π/2. Thus the final solution set is {π/6, π/2, 5π/6}. Neither endpoint is a solution because sin 0 = sin 2π = 0 and the original expression then equals 1.

Each candidate can be checked in the original expression. When sin θ = 1/2, the value is 2(1/4) − 3/2 + 1 = 0. When sin θ = 1, it is 2 − 3 + 1 = 0. These checks confirm membership, while the unit-circle positions and the complete factorisation establish that no other angles in the stated interval are required.

Suppose Tricia corrects the factorisation but keeps the old list {π/6, π/2}. Both listed angles pass substitution. Her check has not detected the missing 5π/6 because membership tests cannot discover an angle that was never listed. The recheck must include a generation audit: for each accepted sine value, have all permitted angles been recovered?

The value sin θ = 1 needs particular attention. The usual two sine forms coincide at π/2 in one cycle. Listing π/2 twice does not create two different solutions. A final set should remove duplicate angle values while preserving genuinely distinct values. The issue is equality of angles in the specified interval, not merely whether their sine values are equal.

Now suppose a copied coefficient had originally led to sin θ = 3/2. Repairing that coefficient may restore a valid range value, but the old statement “no solution” must also be revisited. A range rejection attached to the old equation cannot survive automatically after the equation changes. The new accepted values must be carried all the way through angle generation and interval filtering.

Angle units are part of the mathematical state, not just a calculator preference. A value of 30 in degrees and π/6 in radians describe the same angle, but the number 30 inserted into a radian formula is not the same input. Keep the unit visible when a correction involves a calculator or a change of representation.

A graph can provide a useful independent picture of the solution count, but a rough sketch is not a substitute for exact angle generation. The student should not force an algebraic list to match an imprecise drawing. Use the graph to ask a question—such as whether a symmetric solution may be missing—then establish the answer using the actual trigonometric relation and interval.

Kai Kai’s compact retest record is: “Factored form checked; sine values in range; all angles generated; endpoints tested; duplicates removed.” He does not need to rewrite every word during an examination. The point is to ensure that the edited equation has been connected to a complete interval answer rather than to an old or partial angle list.

This chapter deliberately uses exact familiar angles so the checking logic is visible. With an inverse-trigonometric decimal, the same responsibilities remain. Retain sufficient precision while generating related angles, use the correct unit, filter the full interval and apply final rounding only in the form requested. A rounded principal value is not the whole solution.

The final question is therefore broader than “Does this angle work?” It is: “Does every listed angle work, and does the generation argument account for every angle the question permits?” A post-repair pass needs both sides of that question.

13. Multiple angles, division and lost solutions

Two trigonometric repairs deserve their own treatment: correcting the interval of a transformed angle and restoring solutions lost by division. Both can leave a list of individually valid answers that is nevertheless incomplete. They are therefore ideal tests of whether the learner understands more than substitution checking.

Solve sin 2x = √3/2 for 0 ≤ x < 2π. Put u = 2x. The transformed interval is 0 ≤ u < 4π, not 0 ≤ u < 2π. Over that two-cycle interval, u = π/3, 2π/3, 7π/3 or 8π/3. Dividing each value by 2 gives x = π/6, π/3, 7π/6 or 4π/3.

Each original-variable value passes sin 2x = √3/2. The completeness argument comes from covering both cycles of u before returning to x. A student who solves only one cycle in u will list π/6 and π/3. Those two values pass every membership check, so substitution alone cannot expose the omitted second-cycle solutions.

Suppose Alicia corrects the copied original interval from 0 ≤ x < π to 0 ≤ x < 2π. The equation itself has not changed. The old solutions remain valid, but the required set expands. She should preserve those valid members while reopening interval coverage. This is neither a full algebra restart nor a mere final-format edit.

Endpoints should be tested using the original condition. Here x = 0 gives sine zero, and x = 2π is excluded anyway. If the interval were closed at 2π, that endpoint would still not solve this particular equation. But that coincidence is not a reason to ignore endpoint notation in other questions.

Now solve sin 2x = cos x for 0 ≤ x < 2π. Using sin 2x = 2sin x cos x gives cos x(2sin x − 1) = 0. Therefore cos x = 0 or sin x = 1/2. The first branch gives x = π/2 and 3π/2; the second gives x = π/6 and 5π/6. All four are required.

Dividing the equation 2sin x cos x = cos x by cos x without a separate zero case produces only sin x = 1/2. The division is valid where cos x ≠ 0, but it has discarded the cos x = 0 branch. A repair should restore that branch and validate its values, not merely append two remembered angles without explanation.

At x = π/2, the original equation is sin π = cos(π/2), or 0 = 0. At x = 3π/2, it is sin 3π = cos(3π/2), again 0 = 0. At π/6 and 5π/6, the factorisation shows the equation holds because 2sin x − 1 = 0. The complete zero-product argument accounts for every solution in the interval.

Kai Kai’s old check had tested only the two sine-branch angles. Both passed, so the missing branch remained invisible. His new retest set includes the excluded divisor’s zero cases. This is a discriminating check because it targets the exact information lost by the original division.

A useful contrast is a problem whose original domain already excludes cos x = 0, such as an equation involving tan x. In that context, cancelling a nonzero cos x factor may be legitimate on the stated domain. Do not turn “division can lose solutions” into “division is always forbidden”. The relevant question is whether the divisor is known to be nonzero wherever the original problem is defined.

Post-repair checking also needs to distinguish missing solutions from duplicated representations. Angles differing by 2π may represent the same direction but can be different numerical solutions when the interval includes both. In an interval such as 0 ≤ x ≤ 2π, the values 0 and 2π are distinct inputs. Whether both are solutions depends on the equation and endpoint inclusion, not on an automatic duplicate-removal rule.

For a learner, the practical pattern is to record the original interval, the transformed interval, the candidate families and the return map. When a correction touches any of those four items, recheck the later items. This prevents a repaired coefficient or interval from being attached to an old angle list that no longer covers the question.

14. Derivatives, stationary points and tangents

A derivative can support several different claims: where a function is stationary, what its gradient is at a specified point, whether it is increasing, and how a tangent or normal should be written. Correcting the derivative therefore creates several possible retests. The student should identify which of those claims the actual question uses rather than redoing every calculus idea indiscriminately.

Let f(x) = x³ − 3x² − 9x + 5. Differentiation gives f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1). The stationary x-values are −1 and 3. Substituting into the original function gives f(−1) = −1 − 3 + 9 + 5 = 10 and f(3) = 27 − 27 − 27 + 5 = −22.

The stationary points are therefore (−1, 10) and (3, −22). The second derivative is f″(x) = 6x − 6. At x = −1 it is −12, giving a local maximum; at x = 3 it is 12, giving a local minimum. These are local classifications on the unrestricted cubic, not claims of global extrema over all real x.

Suppose Alicia originally differentiates −9x as +9. She repairs the derivative to 3x² − 6x − 9. The old stationary candidates were derived from another equation and must be regenerated. Once new x-values are obtained, their y-values must be evaluated in f, not f′. The classification then needs to be attached to the corrected points.

The distinction between function and derivative is essential. At a stationary point, f′ is zero, but the point’s y-coordinate need not be zero. A student who records (−1, 0) and (3, 0) has used derivative values as function values. A correct derivative repair does not cure that separate interpretation error.

Now find the tangent at x = 2. The point on the curve is (2, −17), because 8 − 12 − 18 + 5 = −17. The gradient is f′(2) = 12 − 12 − 9 = −9. Point-gradient form gives y + 17 = −9(x − 2), hence y = −9x + 1.

The line passes through the correct point: at x = 2 it gives y = −18 + 1 = −17. Its slope equals the corrected derivative value. Both conditions are needed. A line with slope −9 through the wrong point is not the required tangent, and a line through the right point with the wrong slope is not the required tangent either.

There is an exact structural check available in this example. Subtract the tangent from the curve: f(x) − (−9x + 1) = x³ − 3x² + 4 = (x − 2)²(x + 1). The repeated factor at x = 2 is consistent with equal function value and equal first derivative there. It supplies a useful independent algebraic view of the repaired tangent construction.

Do not overgeneralise that check. A tangent to a polynomial curve can intersect the curve elsewhere; the additional factor x + 1 here gives another intersection at x = −1. That does not make the tangent wrong. The claim is tangency at x = 2, not that the line has exactly one intersection with the entire cubic.

A correction to the constant term would have a different scope. If only +5 were corrected to +7, the derivative would remain unchanged. The stationary x-values and local classifications would remain unchanged, while every function value and tangent intercept would rise by 2. A student should be able to preserve the derivative with a mathematical reason rather than mechanically retest it as though its formula had changed.

The OpenStax discussion of maxima and minima is a useful reference for distinguishing critical points and extrema. In this guide’s examples, the decisive habit is to identify the exact claim required: a candidate stationary x-value, a point, a local classification or a global optimum. These claims require different evidence.

For a corrected derivative, a compact retest set might be: verify the derivative term by term, solve its current zero equation, evaluate the original function at the accepted x-values, classify if asked, and recheck any tangent using both point and slope. The checks should follow the question’s actual outputs. Do not add a tangent calculation to a question that never asks for one.

Tricia’s improvement is to stop treating “the calculus answer” as one object. She separates the derivative, the x-coordinate, the y-coordinate and the interpretation. A repair may affect some of them but not others. That separation makes the recheck smaller and more precise.

15. Integration, constants and signed area

Differentiating an antiderivative is a powerful check, but it cannot reveal every integration error. Two antiderivatives that differ by a constant have the same derivative. A derivative check may therefore confirm the variable-dependent part while leaving an incorrect constant undetected. Bounds and geometric interpretation create further responsibilities.

Suppose F′(x) = 3x² − 6x and F(1) = 5. Integrating gives F(x) = x³ − 3x² + C. Applying the condition gives 1 − 3 + C = 5, so C = 7. The required function is F(x) = x³ − 3x² + 7.

There are two different checks. Differentiating F returns 3x² − 6x. Substituting x = 1 returns 5. The first checks the antiderivative structure; the second checks the constant against the given condition. A candidate F(x) = x³ − 3x² + 5 passes the derivative check but fails the condition because F(1) = 3.

Imagine Kai Kai corrects the power-rule integration but retains C = 5 from an earlier incorrect condition substitution. A post-repair routine that stops after differentiation misses the remaining fault. The recheck must include every condition used to select one member of the antiderivative family.

For the definite integral of 3x² − 6x from 0 to 2, an antiderivative is G(x) = x³ − 3x². The integral is G(2) − G(0) = (8 − 12) − 0 = −4. An arbitrary common constant cancels from this difference, so an omitted constant of integration is not a numerical defect in this definite evaluation.

The negative answer is appropriate for the signed integral because 3x² − 6x = 3x(x − 2) is nonpositive on [0, 2]. If the question instead asks for the geometric area between the curve and the x-axis on that interval, the area is 4. Do not change −4 to 4 merely because “area should be positive” without first checking which quantity was requested.

A more revealing example uses f(x) = x² − 1 on [−2, 2]. The signed integral is [x³/3 − x] from −2 to 2, which equals 4/3. The curve crosses the axis at −1 and 1. It is positive outside [−1, 1] and negative inside that interval.

The geometric area is therefore not the absolute value of the total signed integral. Split at the crossings. On [−2, −1], the area is 4/3. On [−1, 1], the signed integral is −4/3, so the area contribution is 4/3. On [1, 2], the area is 4/3. The total area is 4.

Suppose Alicia corrects an antiderivative but keeps the old “area = 4/3” conclusion. Differentiation and endpoint substitution may both pass, yet she still answers the signed-integral question instead of the geometric-area question. The missing retest concerns interpretation and splitting at sign changes, not the integration formula.

The zero locations are important intermediate results because they determine the splitting intervals. If a coefficient correction changes the roots, the old area partition must be rechecked before evaluating integrals. Conversely, if only a final arithmetic subtraction was wrong and the partition is verified, the student can preserve the sign analysis while recomputing the numerical total.

Check the subtraction order in a definite integral. Upper antiderivative value minus lower antiderivative value is a mathematical operation, not a visual habit. A negative lower value must be subtracted with its sign intact. Writing the two endpoint values separately can make this stage easier to inspect without expanding every simple arithmetic step.

For area between two curves, the same logic becomes top curve minus bottom curve over each interval where their order is fixed. A repaired intersection changes the possible boundaries; a repaired curve equation may change which curve is above. Testing only the final antiderivative cannot establish that the correct region was integrated.

Tricia now asks three questions after an integration repair: does differentiation return the integrand, do the initial conditions or bounds remain correctly applied, and does the resulting signed or unsigned quantity match the question? These questions are different enough that one successful test should not be counted as three.

The result is a practical stopping rule. Once the antiderivative, selecting conditions, interval structure and requested interpretation are established, there is no need to keep recomputing the same integral for reassurance. If a later edit changes the integrand or boundary, reopen the affected checks.

16. Optimisation and feasible global answers

Optimisation combines modelling, algebra, calculus and interpretation. A correction near the start can affect every later stage, while a correction near the end may leave most of the model intact. The recheck should therefore begin with the quantity being optimised and the set of feasible inputs, not with whichever derivative happens to be visible.

As a simple foundation, a rectangle with perimeter 20 has side lengths x and 10 − x. Its area is A(x) = x(10 − x) = 10x − x². For nonnegative side lengths the closed interval is 0 ≤ x ≤ 10; for a nondegenerate rectangle the endpoints are excluded. The area is maximised at x = 5, giving the square with area 25.

Completing the square gives A(x) = 25 − (x − 5)², so A(x) ≤ 25 with equality at x = 5. This proves the global maximum on the feasible interval directly. Differentiation gives the same candidate, but the completed-square argument makes the global bound especially transparent.

Suppose the learner first used the incorrect perimeter relation x + y = 20, then repairs it to x + y = 10. The old objective 20x − x² and its candidate x = 10 no longer belong to the original problem. The model, candidate and reported side lengths need to be regenerated. A derivative check of the old objective would simply confirm correct calculus applied to the wrong model.

Now consider an open box made by cutting equal squares of side x from the corners of a 20 by 12 sheet and folding the sides upward. The base dimensions are 20 − 2x and 12 − 2x, and the height is x. The volume is V(x) = x(20 − 2x)(12 − 2x) = 240x − 64x² + 4x³.

A nondegenerate box requires 0 < x < 6. It is useful to consider the continuous extension to [0, 6], where the endpoint volumes are zero, while remembering that the endpoints are degenerate boxes. Differentiation gives V′(x) = 240 − 128x + 12x² = 4(3x² − 32x + 60).

Setting V′ = 0 gives x = (16 ± 2√19)/3. The smaller value r = (16 − 2√19)/3 is approximately 2.4274 and lies in (0, 6). The larger value is approximately 8.2393 and lies outside the feasible domain. It must not be accepted merely because it is a root of the derivative equation.

To establish the maximum, note that V′ is positive on (0, r) and negative on (r, 6), since the other derivative root lies above 6 and the derivative quadratic opens upward. Thus V increases and then decreases across the feasible interval. The endpoint extension has volume zero, and V(r) is positive. The feasible global maximum occurs at r.

The exact volume is (1792 + 1216√19)/27, approximately 262.6823 cubic units when the sheet dimensions are in the same length unit. The exact cut size should be retained until the requested final rounding. The resulting base dimensions are positive, which provides a geometric feasibility check separate from the calculus.

Imagine Tricia corrects an expansion error in V but leaves the old derivative unchanged. Repair confirmation requires re-expanding the product accurately. Post-repair checking then requires differentiating the corrected polynomial, regenerating stationary candidates, filtering by 0 < x < 6 and establishing the maximum. No single one of those checks substitutes for the others.

Now imagine the polynomial and derivative are both correct, but Kai Kai accepts the larger stationary root. The failure is not calculus technique. It is feasibility. The proper repair is to reject the out-of-domain candidate and complete the comparison or sign argument for the remaining candidate. There is no reason to redo a verified derivative merely because the final interpretation was wrong.

The standard extrema framework distinguishes interior critical points from endpoint considerations on closed intervals. In a word problem, that framework must be combined with the actual feasible domain. A stationary value outside the model’s domain is not a candidate design, however elegant its exact expression may look.

A useful retest of a repaired optimisation solution therefore asks: does the objective represent the intended quantity, do the constraints produce the stated domain, is the derivative current, are all feasible candidates considered, and is the reported result a global answer of the requested kind? Those questions protect the entire chain without insisting that every stage be solved twice.

Do not use “the second derivative is negative” as a universal substitute for the global question. It can establish a local maximum under the relevant conditions, but a global comparison may require endpoints, other candidates or a whole-interval argument. The box example supplies such an argument explicitly rather than relying on a local label alone.

17. Coordinate geometry: equations must agree

Coordinate geometry offers unusually direct post-repair tests because a point often has to satisfy several defining relations at once. A corrected intersection should lie on both objects. A corrected tangent should pass through the point of contact and have the required direction. A repaired distance should correspond to the actual coordinates, not to an old sketch label.

Find the intersections of x² + y² = 25 and y = x + 1. Substituting the line into the circle gives x² + (x + 1)² = 25. Expanding gives 2x² + 2x − 24 = 0, hence x² + x − 12 = 0. Factorisation gives (x + 4)(x − 3) = 0.

The x-values are −4 and 3. Returning to y = x + 1 gives the points (−4, −3) and (3, 4). Both must be checked against both original equations. For (−4, −3), the circle gives 16 + 9 = 25 and the line gives −3 = −4 + 1. For (3, 4), the circle gives 9 + 16 = 25 and the line gives 4 = 3 + 1.

Suppose Alicia corrects an expansion error and obtains x = 3, but retains y = 2 from an old line y = x − 1. The point (3, 2) may look plausible on a rough sketch, yet it fails the current line and circle. Checking the quadratic in x would not reveal the stale y-value. The post-repair check must include the return from x to the full coordinate pair.

Now find the tangent to the circle at (3, 4). The radius from the origin to that point has gradient 4/3. The tangent is perpendicular to the radius, so its gradient is −3/4. Using point-gradient form, y − 4 = −3(x − 3)/4. Multiplying by 4 gives 3x + 4y = 25.

The point test gives 3(3) + 4(4) = 25. The direction test gives (4/3)(−3/4) = −1. These are separate checks. A line through (3, 4) with gradient 4/3 is the radius line, not the tangent; a parallel line with gradient −3/4 but wrong intercept also fails the required tangent claim.

An alternative exact check uses the circle and the proposed line together. Solving y = (25 − 3x)/4 with x² + y² = 25 gives a quadratic with a repeated root at x = 3. That is consistent with the single point of contact for this line and circle. It is a useful algebraic confirmation, although a complete repetition of that calculation is not always necessary once point and perpendicularity are correctly established.

Be careful with vertical and horizontal lines. The gradient-product rule is convenient when both relevant gradients are finite. At a circle point such as (5, 0), the radius is horizontal and the tangent is vertical, x = 5. Trying to force a finite gradient for the vertical tangent would create an unnecessary error. A corrected point can therefore change which form of the direction check is appropriate.

Now take the distance between A = (−1, 2) and B = (3, 5). The coordinate differences are 4 and 3, so the distance is 5. Reversing both differences gives −4 and −3, whose squares produce the same distance. That reversal is not a distance error. It changes the direction of a displacement vector, but not its magnitude.

This distinction matters when reviewing a corrected script. Do not label every changed sign as a confirmed numerical fault. Ask what mathematical object is being computed. For a vector from A to B, the signed differences matter. For the distance, squaring removes their signs legitimately. A good retest evaluates the claim rather than enforcing one visual arrangement of the formula.

If a point correction changes only one coordinate, identify every later use of that coordinate. A gradient, midpoint, distance and line equation may all need attention, but an independent circle radius established elsewhere may not. The Intermediate-Result Criticality guide explains why heavily reused results deserve early checking; the same reuse map selects the post-repair tests.

The most reliable final question in coordinate work is often concrete: “Does this current point or line satisfy the relations that define it?” Return to those relations before trusting a neat coordinate pair. A geometric answer should be supported by geometry as well as by the last arithmetic line.

18. Exactness, rounding and calculator handoffs

A calculator can execute the wrong expression accurately. It can also display a rounded value that looks identical to another value even though the two are not exactly equal. Post-repair checking must therefore identify the current expression, relevant settings, stored inputs and requested precision before treating the display as evidence.

Consider d = √101 − 10. Rationalising gives d = (101 − 100)/(√101 + 10) = 1/(√101 + 10). The exact value is positive and approximately 0.0498756211. Both expressions are mathematically equivalent, but they behave differently if the square root is manually rounded before the subtraction.

If a student rounds √101 to 10.0 first, subtracting 10 gives zero. That cannot be a valid approximation for a later calculation involving 1/d. If the student retains 10.05, the subtraction gives 0.05, which is much better but still not the exact d. The correction should restore an exact or sufficiently precise source before regenerating later numerical results.

The reciprocal provides an exact cross-check: 1/d = √101 + 10, approximately 20.0498756211. Using d = 0.05 gives 20. To three significant figures both may be written as 20.0, while to four significant figures the exact calculation gives 20.05. A matching rounded endpoint can therefore conceal a less accurate intermediate calculation.

The lesson is not that every intermediate decimal is forbidden. It is that a precision policy should match the sensitivity of the later calculation and the required output. Retaining an exact form or calculator-held precision avoids unnecessary information loss. Once the quantity is no longer used downstream, apply the requested final rounding.

Suppose Tricia repairs d from zero to 0.05 and sees a plausible reciprocal. She has removed a dramatic error, but the current check still needs to ask whether the precision is sufficient. The appropriate answer depends on the requested accuracy and how the value is used. There is no universal decimal tolerance that can be applied to every Additional Mathematics problem.

Distinguish exact equality from a numerical residual. Substituting an approximate root into a polynomial may give a small nonzero value. That is expected when the root is rounded. A small residual can support numerical consistency, but its meaning depends on the equation’s scale and behaviour near the candidate. Do not treat “small” as a universal proof that the root is accurate enough.

An exact example avoids that ambiguity. For x = √2 in x² − 2 = 0, symbolic substitution gives zero. Using x = 1.414 gives 1.999396 − 2 = −0.000604. The second result reflects rounding of the candidate, not necessarily an algebraic failure. Replacing a valid exact answer because its rounded display has a residual would be a poor repair.

Calculator provenance can be checked in a short sequence: read the original or corrected written expression; confirm the intended input values; confirm the angle unit if relevant; enter the expression with clear brackets; compare the output with a sign, range or rough magnitude expectation; then record the result with its precision status. These steps are a proposed routine, not a model-specific button guide.

Parentheses deserve particular attention after edits. The expression 1/(x + 2) is not the same as 1/x + 2. Changing a denominator on paper may leave an old calculator entry structurally wrong. A fresh entry from the current written expression is often easier to audit than a series of edits whose history is no longer visible.

Stored variables and previous-answer registers can also remain stale. If k changes from 2 to −2, the written formula may be corrected while the calculator still stores k = 2. Rechecking the final display without checking the stored input repeats the old calculation. The source-result pair must be current together.

Do not add unnecessary precision to the final answer as a substitute for sound checking. A long display is not evidence that the expression was correct. Similarly, do not strip a valid exact answer into a decimal merely because a calculator is available. Follow the required form and preserve the distinction between exact mathematical state and displayed approximation.

Kai Kai uses one sentence during review: “The display belongs to this expression with these inputs.” If he cannot complete that sentence, he recomputes from a known current source. That is a practical way to prevent a repaired page and an unrepaired tool state from producing a falsely reassuring agreement.

19. When correcting one error exposes another

A post-repair check can look worse after a valid correction. This is not a reason to restore a known error. It may mean that an earlier error had been compensating for another, so removing one has made the remaining fault visible. The recheck should follow mathematical validity rather than the temporary closeness of the endpoint to an expected answer.

Let g(x) = −2x² and suppose the question asks for g′(2). The correct derivative is g′(x) = −4x, giving g′(2) = −8. A student incorrectly writes g′(x) = 4x, obtains 8 and then adds an unjustified minus sign because the curve “ought to be going down”. The final number is −8, but the derivative formula and the extra sign operation are both wrong.

Now repair the derivative formula to −4x while leaving the extra sign reversal in the numerical procedure. The calculation gives −8 and then incorrectly changes it to 8. The endpoint has become visibly wrong even though the derivative repair was correct. The remaining problem is the stale compensating sign reversal.

A clean recheck confirms g′(x) = −4x directly from the original function, evaluates it at 2 without an extra sign operation, and obtains −8. A later demand such as g′(1) should also use the corrected derivative directly, giving −4. The purpose of this extra use is to reveal whether the old workaround still survives in the learner’s procedure.

The same issue can occur in scale. Suppose a student’s intermediate length is accidentally doubled while a later formula incorrectly halves it. One endpoint may match the correct result. Correcting the length but retaining the erroneous halving operation exposes the second fault. The response should be to repair the formula, not to double the length again.

However, not every cancellation is an error. Reversing both coordinate differences in a gradient leaves the quotient unchanged and can be entirely valid. Adding the same constant to both endpoint values in a definite integral also cancels legitimately. Before diagnosing hidden faults, establish that each disputed operation is actually invalid for the claim being made.

A second source of false reassurance is shared contamination. Two methods can agree because both began from the same wrongly copied equation. If the original problem is x² − 5x + 6 = 0 but both routes use x² − 5x + 4 = 0, their agreement on roots 1 and 4 does not validate the original question. A repair must return both routes to the correct source.

The corrected original equation factors as (x − 2)(x − 3) = 0. Its roots are 2 and 3. They satisfy the original equation individually, and the factorisation establishes completeness. Repeating two different methods on the wrong polynomial had supplied no protection against the source-copy error. Different-looking working is not automatically independent evidence.

The Error Cancellation and Masking guide explores these hidden faults more broadly. Here the practical trigger is a post-edit result that remains strange, unexpectedly unchanged or worse. Treat masking as one hypothesis, then test the actual steps. Do not assume every surprising result proves a compensation pair.

There is another possibility: the proposed repair itself may be wrong. Confirm it independently before searching for a second fault. The correct discipline is not “never reconsider a correction”; it is “do not reverse a verified correction merely to recover a familiar endpoint”. Evidence can legitimately reopen the correction if it shows the new line is invalid.

After finding the second fault, recheck the local chain from the most recent trustworthy state. Identify which downstream results used either of the two errors. A result that happened to be numerically correct may still need a fresh derivation if its previous justification depended on invalid working. Keep valid mathematics, not unsupported coincidences.

For a learner, this is a useful emotional distinction. The visible answer may move away from the expected number while the solution becomes logically better. The aim is not to make every intermediate moment look successful. It is to finish with a route whose correctness does not depend on two mistakes cancelling.

20. Build a small but useful retest set

A useful retest set is a small collection of checks chosen for the actual change. It is not a universal ten-item ritual. If a final unit label is corrected, the set may be tiny. If a denominator, parameter or branch structure is corrected, the set may need to include domain, completeness and several downstream uses.

Start with the claims the solution must make. For a repaired tangent problem, the claims might be: the function is copied correctly, the derivative is current, the point lies on the curve, the slope is correct and the line has that slope through that point. A test set should cover these claims with clear evidence. Repeating a point substitution five times does not cover the slope claim.

Changed object	Confirm the repair	Recheck its consequences
Expanded expression	Expand from the original factors	Regenerate roots or values that used it
Domain or interval	Read the original restriction	Refilter candidates and test boundaries
Derivative	Differentiate the current function	Recheck stationary points, slopes and requested classifications
Antiderivative	Differentiate back	Reapply conditions, bounds and area interpretation
Calculator input	Compare with the current written expression	Regenerate numerical outputs and final rounding

The table is a planning aid, not a completeness theorem. A particular problem may require a check not listed here. For instance, a parameter correction can change the degree of an equation, while a geometric correction can create a vertical line. The learner must still inspect the actual mathematics.

Prefer tests with different purposes. One can target the known old fault, another the first important downstream use, and another a boundary or completeness condition that the old method might have missed. The set should not merely contain several ways of repeating the same assumption.

For the repaired radical equation in Chapter 7, a useful set includes the corrected quadratic, the original sign restriction and original-equation validation of the candidates. For the repaired integration example, differentiation alone is insufficient, so the selecting condition F(1) = 5 belongs in the set. These choices follow the problem’s structure, not a fixed number of checks.

Ask whether a test is still current after a later edit. A check performed on yesterday’s formula or on an earlier version of the same script does not automatically apply to the new version. If the changed quantity lies outside the test’s dependencies and a clear argument establishes that, the evidence may remain valid. Otherwise refresh the check.

Also ask whether the expected result is trustworthy. A mark-scheme answer, an official given result, an independently established identity and a student’s rough guess have different evidential roles. Do not silently use a guess as the standard against which a repaired solution is judged. In this guide’s original exercises, the worked solutions show how expected results are obtained.

A pass should be described by scope. “The line passes through the point” is precise. “The tangent is correct” requires the slope or tangency relation too. “Both roots satisfy the polynomial” is precise. “These are all permitted solutions” additionally requires completeness and domain reasoning. Scope-labelled passes prevent small checks from being promoted into larger claims.

There is no honest general conversion from “five checks passed” to “the answer is 99 per cent correct”. Checks may share assumptions, miss the same fault or apply only to part of the claim. Treat the evidence qualitatively unless a real probabilistic model has been defined and justified. In school mathematics, exact reasoning is usually more useful than an invented confidence percentage.

Stop adding tests when the required mathematical claims are established and further checks would simply repeat the same evidence at disproportionate cost. That does not mean certainty is based on a feeling. It means the student can point to the derivation, conditions, validation and output form that support the conclusion.

Reopen the set when another material edit occurs, a candidate changes, a previously omitted case is found or a contradiction appears. A checking plan is tied to a current solution, not a permanent badge. This is the point at which post-repair testing becomes a disciplined continuation habit rather than a box-ticking ceremony.

The useful balance is clear: do not trust a repaired result merely because it looks better, and do not demand a complete second solution for every small edit. Choose checks that have a reason to be there, know what each establishes, and stop when the current answer has the evidence it actually needs.

21. Use the method under examination time pressure

During a timed paper, the whole framework should shrink into a few mathematical decisions. You are not being asked to draw a software-testing diagram beside every correction. You are trying to avoid two expensive mistakes: continuing from a result that has not been properly repaired, and spending the remaining time redoing work that the correction did not affect.

Begin with the edit’s consequences rather than its appearance. Replacing a plus sign with a minus sign takes little ink, but it may change a derivative, its roots and a tangent. Rewriting an entire final sentence may change no mathematics at all. The amount of checking should follow the first kind of consequence, not the size of the crossed-out region.

A useful compressed sequence is source, consequence, condition, answer. Check the corrected source against the original problem. Follow its first important downstream use and any distinct later uses that can change. Restore the relevant domain, branch or completeness condition. Finally make sure the current result answers the actual question. This sequence is a proposed working aid, not an official marking convention.

Suppose you change a quadratic’s constant term and have very little time. Merely ticking the new constant does not establish the old roots. Expand the current factorisation or solve the current quadratic, then test the resulting candidates against the original equation. A previous check of the old roots has expired for this purpose because it tested a different polynomial.

Now suppose the correction is only from “3.14” to “π” in a final exact-answer line and all earlier work already used π. The symbolic derivation may remain unchanged. Confirm that the requested answer is exact and that no later calculation used 3.14. There is no mathematical reason to repeat the entire route just because the final presentation was corrected.

Time pressure does not make a false statement acceptable, but it can limit what can be completed. When the recheck remains unresolved, leave a clear current equation and the justified progress you actually have. Do not manufacture a final candidate by choosing the number that looks most familiar. A visible unresolved calculation is more honest than a confident answer produced by an unsupported edit.

For linked questions, look at whether the next part depends on your own result or on a result explicitly supplied in the question. These are different starting points. If a later part explicitly permits using a stated result, your failed attempt to derive it need not stop you using that given result there. Read the wording carefully; do not invent a general rule that every “hence” question supplies permission to bypass the earlier derivation.

Keep mathematical validity separate from guesses about marks. This guide does not assign method marks to its examples, and it does not promise what a marker will award for a particular crossed-out line. Those decisions belong to the qualification and question. The practical aim is to make the surviving mathematical argument readable and correctly connected to the problem’s conditions.

If you pause the question, leave a useful restart note. “Current k = −2; tangent not yet rechecked” is enough to distinguish the repaired parameter from the unfinished consequence. On return, read that note and the original condition before continuing. Do not rely on a vague memory that “the sign was fixed somewhere”.

Do not change a correct answer solely because repeated checking has made it feel unfamiliar. Ask for a specific reason: a failed substitution, an incompatible condition, a copied coefficient, a missing case or an invalid inference. If no new evidence appears and the required claims are already established, stop. Repetition without a new checking purpose can consume time without answering a new mathematical question.

During practice, measure the routine with comparable tasks rather than one dramatic example. Note whether you selected the right retest, caught stale dependents and stopped after adequate evidence. A shorter checking time is useful only if the required claims remain supported. It is not progress to become faster by omitting the domain or completeness stage.

One practice exercise can deliberately allocate a small checking window after a worked correction. Give the learner a repaired derivative, one stale coordinate and an independent correct distance. Ask them to choose the next two actions. The best response identifies the coordinate’s dependence and preserves the distance. It does not waste the window checking the independent distance simply because that calculation is easier.

The fuller Checkpoint Architecture guide addresses where to check a long solution. Here, the trigger is narrower: a material edit has occurred. Use the changed line to select a limited, purposeful set of checks, then return to solving once the current result is justified.

The examination-day stopping question is concrete: “Can I now explain why this corrected result and the conclusions I am using from it satisfy the original task?” If yes, continue. If no, identify the missing evidence. That is more useful than waiting to feel absolutely reassured or trying to turn a rushed correction into a ceremonial second solution.

22. An integrated quadratic correction case

This case brings the separate checks together. It is an original teaching problem, not an examination-board question. Let f(x) = x² + ax + b, where the graph passes through (1, 0) and (3, 0). Find a and b, the vertex, the tangent at x = 4, and the geometric area between the curve and the x-axis from x = 1 to x = 3.

The two point conditions give 1 + a + b = 0 and 9 + 3a + b = 0. Subtracting the first equation from the second gives 8 + 2a = 0, so a = −4. Then 1 − 4 + b = 0 gives b = 3. Thus f(x) = x² − 4x + 3 = (x − 1)(x − 3).

Those parameter values should be checked against both original points. We obtain f(1) = 1 − 4 + 3 = 0 and f(3) = 9 − 12 + 3 = 0. This confirms the two defining conditions. Checking only one point would leave one equation of the parameter system untested.

Now examine a deliberately faulty script. Alicia correctly finds a = −4 but writes b = 5. She then correctly completes the square for that wrong function as x² − 4x + 5 = (x − 2)² + 1. She writes “vertex (2, 1)” and “no real x-intercepts”. Elsewhere she independently differentiates −4x incorrectly as +4, obtaining f′(x) = 2x + 4.

Using the wrong constant and wrong derivative, she finds f(4) = 5 and a slope of 12. Her old tangent is y − 5 = 12(x − 4), or y = 12x − 43. Finally she integrates the wrong function from 1 to 3 and obtains 8/3, presenting that as the required geometric area.

The script contains two distinct source faults: the constant b and the derivative’s linear-term sign. They interact in the tangent calculation, but one does not cause the other. Correcting b cannot justify retaining or changing the derivative by intuition. The derivative must be checked against the current function, and that check reveals the separate sign error.

First correction: restore the constant

Replace b = 5 with b = 3 using the original point equations. The constant’s repair is confirmed by f(1) = f(3) = 0. The old vertex y-value, root claim, tangent point and area calculation all used b = 5, so each needs fresh evidence. The stationary x-coordinate may survive, but only because the constant does not affect the derivative or the axis of symmetry.

Completing the square for the corrected function gives f(x) = (x − 2)² − 1. The vertex is therefore (2, −1). The x-coordinate is unchanged, while the y-coordinate decreases by 2. This is a precise preservation decision: keep the axis x = 2, not the entire old point.

The old statement “no real roots” must be retired. Factorisation gives roots 1 and 3, which are also the given intercepts. Both original point checks pass. Because the leading coefficient is 1 and the complete quadratic factorisation is displayed, the root set is complete. One successful substitution would not have supplied that last conclusion on its own.

Second correction: restore the derivative

The derivative of the corrected function is f′(x) = 2x − 4. The constant’s correction did not change this derivative; the earlier expression 2x + 4 was independently wrong. A student should not say “I changed the derivative because b changed.” The justification is the derivative of −4x, which is −4.

The stationary condition 2x − 4 = 0 gives x = 2. The second derivative is 2, which is positive. Together with the completed-square form, this establishes the minimum at (2, −1). The completed square also establishes that it is a global minimum over all real x, not merely a local classification at one point.

At x = 4, the current function value is f(4) = 16 − 16 + 3 = 3. The current gradient is f′(4) = 8 − 4 = 4. Both old tangent inputs have changed: the point was based on b = 5 and the gradient on the wrong derivative. Rechecking only the gradient would leave the wrong point alive.

Rebuild and test the tangent

Using the current point and slope, the tangent is y − 3 = 4(x − 4), hence y = 4x − 13. At x = 4 it gives y = 3, so the line passes through the correct point. Its gradient is 4, agreeing with the derivative at that point. These two conditions establish the tangent for this differentiable curve.

There is a particularly efficient independent algebraic check. Subtract the line from the curve: f(x) − (4x − 13) = x² − 8x + 16 = (x − 4)². The difference is zero at x = 4 and has derivative zero there. It is positive elsewhere, so the line touches the parabola at that point and lies below it.

This check would reject the old line y = 12x − 43. At x = 4 the old line gives 5 rather than the current f(4) = 3. Even before testing direction, the point condition fails. A tick on the old tangent’s own point-gradient rearrangement would only show that the wrong inputs were rearranged consistently.

Suppose the script also contains an independent calculation of the distance between (0, 0) and (3, 4). Its value 5 does not use a, b, f or f′. Preserve it. It is included in this teaching case to make selective checking observable: the learner should explain why the distance is outside the correction’s mathematical scope.

Rebuild and test the area

On the interval [1, 3], the corrected factorisation (x − 1)(x − 3) is nonpositive. The curve lies on or below the x-axis. Therefore the geometric area is the integral of −f(x), not the signed integral of f(x). The old positive-integrand interpretation came from the wrong function and must be reconsidered.

An antiderivative of f is H(x) = x³/3 − 2x² + 3x. Its endpoint values are H(3) = 9 − 18 + 9 = 0 and H(1) = 1/3 − 2 + 3 = 4/3. The signed integral is H(3) − H(1) = −4/3. The geometric area is therefore 4/3 square units.

Differentiate H to check the integration: H′(x) = x² − 4x + 3. Then check the interval sign and the subtraction order. These are distinct responsibilities. A correct antiderivative could still be paired with reversed limits or an incorrect area interpretation.

A second exact route uses u = x − 2. The interval becomes −1 ≤ u ≤ 1, and f(x) = u² − 1. The area is therefore the integral of 1 − u² from −1 to 1. Its value is [u − u³/3] from −1 to 1, giving 2/3 − (−2/3) = 4/3.

The agreement is useful because the second route handles the symmetric region in a different form. Still, both routes depend on the corrected function and interval. Their agreement cannot substitute for checking those shared inputs against the original problem. Independence is never improved by forgetting the assumptions the routes actually share.

What the complete recheck establishes

The current answer is now a = −4, b = 3; vertex (2, −1); tangent y = 4x − 13; geometric area 4/3. The parameter equations, vertex representation, tangent conditions and area interpretation each have visible support. The result is more than a set of plausible final numbers.

The recheck also records what survived. The axis x = 2 remained valid because the constant does not change it. The independent distance remained 5 because it used unrelated coordinates. The original point conditions remained authoritative throughout because they came from the question, not from the faulty script.

What did not survive was equally clear: b = 5, vertex (2, 1), the no-real-roots claim, the derivative 2x + 4, the tangent inputs (4, 5) and 12, the old tangent, and the old area conclusion. Each was either corrected or retired with a reason. None remained active merely because it was boxed before the first edit.

For a teacher, the strongest assessment question is not “Did you get 4/3?” Ask: “Why did the axis stay the same while the area changed, and why did the derivative need a separate correction?” A learner who can answer has shown a useful understanding of dependence, invariance and independent faults. That understanding is the point of the case.

23. Ten diagnostic exercises with explained answers

These original exercises assess a decision that ordinary answer checking can overlook: what should be checked after a specific correction? Read the task and the described repair, choose the next useful check, and explain what that check establishes before opening the worked answer. All variables and solutions are real unless an interval states otherwise. These are teaching exercises, not official examination questions or a validated diagnostic instrument.

Do not award yourself full credit simply for finding the numerical answer. For each exercise, identify the changed object, one result that needs fresh evidence, and the reason your check can reveal the remaining defect. When appropriate, identify work that can be preserved. This makes the difference between repairing a calculation and understanding the consequences of its repair visible.

Exercise 1: the denominator that disappeared

Task. Solve (x² − 4)/(x − 2) = 4. A learner repairs the numerator factorisation to (x − 2)(x + 2), cancels the common factor and writes x = 2. They then check that 2 + 2 = 4. Is that enough? Give the correct solution set and the smallest decisive additional check.

Open the worked answer and the checking decision

The original expression excludes x = 2 because its denominator would be zero. On the domain x ≠ 2, the equation simplifies to x + 2 = 4. Its only algebraic candidate is x = 2, but that candidate is outside the original domain. Consequently the original equation has no real solution.

The decisive check is not another evaluation of x + 2. It is a return to the original denominator. The student’s current check would also pass for the very value at which the original expression is undefined. It therefore tests the extended linear expression, not the original rational equation. The factorisation is correctly repaired while the candidate’s acceptance remains invalid.

There is no need to start the factorisation again. Preserve the valid identity x² − 4 = (x − 2)(x + 2). Restore its condition of use in the quotient, then reject the candidate. Completeness follows because every original solution would have to satisfy the equivalent reduced linear equation on the permitted domain, and its only candidate has been excluded.

A common unhelpful response is to write that the answer is infinity. The equation has not asked for a limit, and division by zero has not produced a valid real value. Another unhelpful response is to cross out the whole method because cancellation supposedly caused an error. Cancellation was legitimate away from x = 2; losing the restriction was the fault.

Contrast test. Change the right side to 5 during practice. The reduced equation now gives x = 3, which is allowed, and the original quotient is (9 − 4)/(3 − 2) = 5. Explain why the same factorisation and cancellation lead to a valid solution in this variant but no solution in the original. A learner who can explain the contrast has more evidence than someone who merely remembers to reject the number 2.

Exercise 2: a valid root does not define every later expression

Task. Solve √(2x + 3) = x. A repaired quadratic gives candidates 3 and −1. The learner accepts both and then calculates y = 1/(x − 3) using the positive candidate. Audit the root acceptance and the later calculation separately.

Open the worked answer and the checking decision

The principal square root is nonnegative, so an original solution must have x ≥ 0. Squaring gives x² − 2x − 3 = 0, or (x − 3)(x + 1) = 0. At x = 3 the original equation reads √9 = 3 and passes. At x = −1 it reads √1 = −1 and fails. The original equation’s solution set is therefore {3}.

The later expression y = 1/(x − 3) is undefined at that valid root. This does not make x = 3 an invalid solution of the radical equation. It means the additional expression has a condition that the selected root does not meet. Keep the two claims separate: root membership passes; the proposed downstream reciprocal cannot be evaluated as a real number.

A student can therefore repair the candidate list successfully and still need to stop a later calculation. That is exactly what a downstream recheck is meant to reveal. It would be wrong to choose −1 merely because its reciprocal is defined, since −1 fails the original equation. It would also be wrong to replace the zero denominator with a small invented number to make the calculator continue.

If a complete task demanded both the radical equation and a real value of this reciprocal at the same x, the combined requirements would have no admissible solution. If the task only asks for the radical equation, answer x = 3. The difference comes from the scope of the stated requirements, not from a preference for one interpretation.

Contrast test. Replace the later expression with y = 1/(x + 3). The same accepted root gives y = 1/6. The radical solution did not need to be regenerated; only the downstream denominator condition and evaluation changed. State which work survives and why. This checks whether you can carry a valid source into a new use without assuming that every new use is automatically allowed.

Exercise 3: a combined logarithm hides the original restrictions

Task. Solve log₂(x − 2) + log₂(x + 2) = 5. A learner repairs (x − 2)(x + 2) to x² − 4, obtains x = ±6, and checks log₂(36 − 4) = 5 for both candidates. What has been checked, and what has not?

Open the worked answer and the checking decision

The original real logarithms require x − 2 > 0 and x + 2 > 0, giving x > 2. On that domain, combining them gives log₂(x² − 4) = 5, hence x² − 4 = 32 and x² = 36. The candidates are −6 and 6, but only 6 satisfies the original domain.

At x = 6, the original arguments are 4 and 8. Their base-2 logarithms are 2 and 3, so the original sum is 5. At x = −6, the original arguments are −8 and −4; the two real logarithms are not defined. The final solution set is {6}.

The learner’s check confirms the combined expression at both candidates. It does not confirm that both original logarithms exist. A positive product is not the same as two positive factors: two negative factors also have a positive product. Forgetting the domain can therefore make a correct algebraic product look like permission to accept an invalid original input.

Preserve the repaired difference-of-squares identity and the resulting candidate equation. Refilter the candidates using the original arguments. There is no need for a new quadratic method, a decimal approximation, or a graphing tool to settle the question. The missing evidence is a sign condition available directly from the original statement.

Contrast test. Consider the different equation log₂(x² − 4) = 5 as the original problem. Its domain is x² − 4 > 0, so both ±6 are allowed. The two original problems share a transformed expression but do not share a domain. A satisfactory explanation identifies this difference rather than saying that negative x-values are always forbidden in logarithm questions. Logarithm arguments, not the variable’s sign in isolation, control the real domain.

Exercise 4: solving a branch is not accepting its candidate

Task. Solve |x − 2| = 2x + 1. A learner repairs the negative branch to 2 − x = 2x + 1 and obtains x = 1/3. The positive branch gives x = −3. They present both values because both linear equations were solved correctly. Choose the next checks.

Open the worked answer and the checking decision

In the branch x ≥ 2, the equation is x − 2 = 2x + 1, giving x = −3. This candidate does not satisfy x ≥ 2 and must be rejected. It also makes the original right side negative, while an absolute value is nonnegative. Correctly solving the branch equation has not established that the branch applies at the resulting value.

In the branch x < 2, the equation is 2 − x = 2x + 1, giving x = 1/3. It satisfies the local branch condition. The original equation gives |1/3 − 2| = 5/3 and 2/3 + 1 = 5/3, so membership is confirmed. The solution set is {1/3}.

The global necessary condition 2x + 1 ≥ 0 gives x ≥ −1/2. It supplies another quick rejection of −3, but it does not itself prove that 1/3 solves the equation. Use necessary conditions as filters, not as replacements for all other requirements. The original substitution completes the membership check.

Completeness comes from the two cases covering all real numbers, with the boundary x = 2 assigned to the first case. At that boundary the original sides are 0 and 5, so it is not a solution. No third branch is missing. The negative-branch repair can be confirmed without re-solving a correctly stated positive-branch equation, but its candidate still needs its condition checked.

Contrast test. Solve |x − 2| = 1. Here both branch candidates are valid: x = 1 and x = 3. Explain why the existence of an absolute value does not imply that every such equation has two solutions, and why one rejected branch in the original problem is a mathematical outcome rather than evidence that the branching method failed.

Exercise 5: a parameter correction removes the equation’s variable

Task. Consider (a − 2)x² + (a − 2)x − 2 = 0. A parameter value is corrected from a = 3 to a = 2. The learner retains the old roots 1 and −2 and tries to re-enter the quadratic formula. What is the correct first recheck?

Open the worked answer and the checking decision

Substitute the corrected parameter into the original equation before selecting a method. At a = 2, both variable coefficients vanish and the equation becomes −2 = 0. That statement is false for every real x, so the equation has no solution. It is not a quadratic and not even a nonconstant linear equation.

At the old a = 3, the equation was x² + x − 2 = 0, giving roots 1 and −2. Those roots were correct for the old parameter but do not survive the correction. Testing either in the corrected original equation gives −2 rather than zero. The old root set and the old formula application must be retired.

This is a structural change, not a difficulty with calculator entry. The quadratic formula divides by twice the leading coefficient, which is now zero. An error message is not the explanation of the mathematics; the original substituted equation explains why the formula is inappropriate and why no x can work.

Do not conclude that every x is a solution merely because all x-terms disappeared. That conclusion would apply to an identity such as 0 = 0, not a contradiction such as −2 = 0. The constant part must still be examined. This distinction is a useful boundary test for any general parameter procedure.

Contrast test. Replace the original constant −2 by a − 2. At a = 2 the whole expression becomes zero and the equation is 0 = 0, so all real x satisfy it. For a ≠ 2, division by a − 2 gives x² + x + 1 = 0, which has no real roots. The parameter regimes reverse the answer dramatically. State the case distinction explicitly instead of extending a formula across a point where its derivation divides by zero.

Exercise 6: the final variable and the stated domain both matter

Task. Solve x⁴ − 10x² + 9 = 0 subject to x < 0. After correcting an arithmetic error, a learner obtains u = 1 and u = 9 using u = x². They report x = 1 and x = 3. Audit the two handoffs from the quadratic to the final answer.

Open the worked answer and the checking decision

The transformed equation is u² − 10u + 9 = (u − 1)(u − 9) = 0. Both u-values are nonnegative and therefore belong to the range of x². Returning to real x gives x = ±1 and x = ±3. That is the complete real root set before the additional restriction x < 0 is applied.

The requested solution set is {−3, −1}. For either magnitude, the original quartic vanishes: 1 − 10 + 9 = 0 at ±1, and 81 − 90 + 9 = 0 at ±3. But positive roots do not meet the question’s domain. The learner omitted the negative preimages and then retained precisely the values the question excludes.

The repaired quadratic calculation can be preserved. What requires correction is the return from u to x and the final domain filter. Re-solving the quadratic by the formula rather than factorisation would not address either fault. A different method applied to the already correct stage would consume time without testing the missing claims.

Checking only that 1² = 1 and 3² = 9 establishes two preimages, not all preimages, and it says nothing about x < 0. Checking only that −1 and −3 are negative would establish the domain condition but not the quartic equation. The complete response connects generation, return, filtering and original membership.

Contrast test. Restrict the same equation to −2 < x < 0. Only −1 remains. There is no need to regenerate all four algebraic roots because the polynomial has not changed; refilter the verified candidates. Explain why a restriction edit and a coefficient edit require different rechecks. This tests selective preservation rather than the mechanical repetition of every stage.

Exercise 7: equal directions can be different permitted inputs

Task. Solve cos 2x = cos x for 0 ≤ x ≤ 2π. A learner corrects the double-angle identity, obtains cos x = 1 or cos x = −1/2, and lists 0, 2π/3 and 4π/3. They omit 2π because it points in the same direction as zero. Is the list complete?

Open the worked answer and the checking decision

Using cos 2x = 2cos²x − 1 gives 2cos²x − cos x − 1 = 0. It factors as (2cos x + 1)(cos x − 1) = 0. The accepted cosine values are −1/2 and 1. Over the closed interval, cos x = 1 gives both x = 0 and x = 2π.

The other value gives x = 2π/3 and x = 4π/3. Thus the complete answer is {0, 2π/3, 4π/3, 2π}. The endpoints are different real inputs even though they describe the same direction after one full turn. Both belong to the stated closed interval, and both satisfy the original equation because their relevant cosine values are 1.

Every angle in the learner’s shorter list passes substitution. That does not prove completeness. The missing check is endpoint coverage under the actual interval, not another evaluation of the already listed interior angles. It is possible to have perfect membership checking and an incomplete answer.

At x = 2π/3, cos 2x = cos(4π/3) = −1/2, matching cos x. At x = 4π/3, cos 2x = cos(8π/3) = −1/2. These original checks confirm the interior members. The complete factorisation and full interval generation establish that no additional families have been omitted.

Contrast test. Change the interval to 0 ≤ x < 2π. The shorter three-value list is now correct. Only the endpoint status changed. Explain why removing 2π in this variant is required, while removing it from the original problem is an error. The correct procedure reads the interval; it does not apply a universal rule that equivalent directions must always be collapsed.

Exercise 8: a tangent can cross a curve

Task. Let f(x) = x³ − 6x² + 9x + 2. A learner repairs the derivative to 3x² − 12x + 9 but leaves the stationary points’ y-values equal to zero. They also reject a tangent at x = 2 because it crosses the curve. Find the required corrected points and tangent, then identify the mistaken checks.

Open the worked answer and the checking decision

The derivative factors as 3(x − 1)(x − 3), so the stationary x-values are 1 and 3. Function values must come from f, not f′: f(1) = 1 − 6 + 9 + 2 = 6 and f(3) = 27 − 54 + 27 + 2 = 2. The stationary points are (1, 6) and (3, 2), not points on the x-axis.

The second derivative is 6x − 12. It is −6 at x = 1 and 6 at x = 3, giving a local maximum at the first point and a local minimum at the second. These checks attach classifications to the corrected coordinates. Neither derivative value supplies the point’s y-coordinate.

At x = 2, f(2) = 8 − 24 + 18 + 2 = 4 and f′(2) = 12 − 24 + 9 = −3. The tangent is y − 4 = −3(x − 2), or y = −3x + 10. It passes through (2, 4) and has the derivative’s gradient there. These are the relevant tangent conditions.

The curve minus the tangent is x³ − 6x² + 12x − 8 = (x − 2)³. This difference changes sign at x = 2, so the curve crosses its tangent. That is fully consistent with tangency at this point. The learner’s rule that a tangent never crosses a curve is not a valid general test for differentiable curves.

The post-repair plan should preserve the current derivative after confirming it, regenerate the original-function coordinates, and test the tangent using point and slope. Do not change a correct tangent to satisfy a false geometric rule. A checking routine can itself contain a misconception; when a sound calculation fails an invalid test, repair the test rather than the mathematics.

Exercise 9: three integration claims need three kinds of evidence

Task. Given F′(x) = 4x − 6 and F(2) = 1, a learner corrects the variable-dependent terms to F(x) = 2x² − 6x + C but retains C = 1. They then claim that the geometric area between y = 4x − 6 and the axis on [0, 3] is zero. Their only check is differentiating F. Audit the claims.

Open the worked answer and the checking decision

Differentiation confirms that 2x² − 6x + C has derivative 4x − 6 for any constant C. The selecting condition gives F(2) = 8 − 12 + C = 1, so C = 5. The required function is F(x) = 2x² − 6x + 5. The old C = 1 would give F(2) = −3 and therefore fails the condition.

The signed integral of 4x − 6 from 0 to 3 is [2x² − 6x] from 0 to 3, which equals zero. The geometric area is different. The line crosses the axis at x = 3/2. It lies below the axis before that point and above afterwards, so the positive and negative signed contributions cancel but the areas add.

Each triangular region has base 3/2 and height 6, hence area (1/2)(3/2)(6) = 9/2. The total geometric area is 9. The signed integral being zero is correct; presenting it as the geometric area is not. A correction should preserve the valid signed calculation while repairing the interpretation and the required area calculation.

There are three distinct claims: an antiderivative family, the particular function meeting F(2) = 1, and an unsigned area over a specified interval. Differentiation checks the first. Substitution in the selecting condition checks the second. Sign analysis and appropriate integration or geometry check the third.

Contrast test. Restrict the geometric-area question to [3/2, 3]. The integrand is nonnegative throughout, so its signed integral and geometric area both equal 9/2. Explain the mathematical condition that makes the two quantities coincide there. The lesson is not that an integral is never an area, but that its relation to area depends on the function’s sign over the interval.

Exercise 10: a point check cannot certify a tangent’s direction

Task. A circle has equation x² + y² = 10. Its intersection with y = 3x in the first quadrant is corrected to P = (1, 3). The learner keeps the line y = 3x as the tangent because it passes through P. Construct a decisive retest and give the tangent.

Open the worked answer and the checking decision

The point is valid: 1² + 3² = 10 and 3 = 3(1). It is an intersection of the stated circle and line. But the line y = 3x passes through the centre of the circle, so at P it has the radius’s direction rather than the tangent’s direction. Point membership is necessary for the tangent but does not determine its slope.

The radius gradient is 3. A perpendicular tangent has gradient −1/3. Using P, the tangent is y − 3 = −(x − 1)/3, or x + 3y = 10. The point check gives 1 + 9 = 10; the gradient check gives 3(−1/3) = −1.

An independent algebraic test substitutes y = (10 − x)/3 into the circle. Multiplication by 9 gives 9x² + (10 − x)² = 90. Expanding and collecting gives 10x² − 20x + 10 = 10(x − 1)² = 0. The repeated root identifies the contact at x = 1, with y = 3.

Only one of these complete approaches is needed for an ordinary solution, but comparing them teaches what each condition establishes. The old point-only check would also pass for infinitely many other lines through P. A decisive test must distinguish the tangent from those alternatives. Perpendicularity or the repeated-intersection calculation does that.

After the point correction, preserve the circle equation and the original intersection line as the problem’s givens. Retire the claim that the intersection line is also the tangent. Rebuild the tangent from the current point and its correct direction. This final exercise tests whether the learner can preserve valid objects while rejecting an invalid relationship between them.

How to read your diagnostic result

A numerical total alone hides important differences. A learner may solve every equation but repeatedly miss which old conclusion remains stale. Another may select the right recheck but make a fresh arithmetic error executing it. Record those separately. The first calls for work on dependency and evidence scope; the second may call for practice with the particular mathematical operation.

For each item, write one sentence beginning “My check establishes…” and a second beginning “It does not establish…”. For Exercise 7, substitution establishes membership of the listed angles but not completeness. For Exercise 9, differentiation establishes the antiderivative family but not the selecting constant. These statements reveal whether a passed check is being used to justify more than it can support.

Do not infer a student’s general ability or emotional state from one response. Use the observed working to choose a small next teaching action. Revisit a genuinely different problem with the same checking demand later. The purpose of this laboratory is to make the next lesson more precise, not to attach a permanent label to the learner.

24. A practical lesson and learner-specific routes

The following lesson is a proposed teaching design for learners who already know the mathematics in the selected examples. Its purpose is to make a post-correction decision observable: can the learner identify what has changed, select a useful check and connect the repaired result to the surviving solution? It is not a test of how many chapter names the learner can remember.

Use a ninety-minute session as one possible arrangement, not a prescribed optimum. The suggested allocation is eight minutes for an independent entry task, twelve for discussion, fifteen for modelling, twenty for paired or small-group practice, fifteen for a mixed individual task, fifteen for feedback and a changed problem, and five for an exit record. Adapt the allocation when actual working shows that a prerequisite needs attention.

The general design is informed by the Institute of Education Sciences practice guide, Organizing Instruction and Study to Improve Student Learning, which recommends alternating worked examples with problem solving, asking explanatory questions and distributing learning over time. Those recommendations support broad instructional choices. They do not validate this particular ninety-minute schedule or establish that this named checking routine produces a specified examination gain.

Before the lesson: select the mathematical demand

Choose one mathematical family before combining several. For a learner beginning this work, use factorisation and original-equation substitution. For someone already secure there, use a domain-sensitive rational equation, a trigonometric completeness problem or a derivative with several downstream uses. The selected mathematics should be familiar enough that the checking decision, rather than an untaught technique, is the main difficulty.

Prepare one correct worked solution, one deliberately faulty version and one changed problem. Write down the intended defect before constructing the faulty version. For instance, a correct factorisation followed by an invalid domain acceptance tests a different skill from an incorrect factorisation followed by otherwise consistent arithmetic. Do not mix those versions accidentally and then interpret the student’s confusion as a diagnostic finding.

Independently solve every teaching item. A plausible answer key is not enough. Check domains, interval endpoints and the interpretation of any requested area or extremum. When a learner disputes the supplied answer, inspect the mathematics rather than treating disagreement itself as evidence of weakness. The teacher’s worked solution is also a solution that can contain errors.

Minutes 0–8: collect an independent starting sample

Present the entry task without the full checklist visible. One suitable prompt is: “The function was copied as x² − 6x + 9, but the original question says x² − 6x + 8. The constant has been corrected. Which of the old derivative, vertex, roots and tangent conclusions can be kept, and which need checking again?” Require reasons rather than a sequence of ticks.

The derivative 2x − 6 is unaffected by the constant correction. The vertex’s x-coordinate remains 3, but its y-coordinate changes from zero to −1. The roots change from the repeated root 3 to 2 and 4. A tangent at a specified x keeps its slope but changes its vertical position. These distinctions make selective preservation visible without introducing unfamiliar calculus.

Do not provide these answers while collecting the sample. Ask the learner to make the decisions first. Record whether each decision was independent, prompted or supplied. A correct response after a detailed hint has value as guided practice, but it should not be recorded as independent success. Keeping that distinction makes later comparison more honest.

Minutes 8–20: discuss the decision, not just the answer

Ask the learner to point to the actual use that connects the corrected constant with a changed conclusion. “The vertex changes” is less precise than “its y-coordinate uses the function value, while its x-coordinate is still determined by 2x − 6 = 0”. A short explanation should identify the mechanism rather than repeat the instruction to check carefully.

Next ask what could remain wrong even after the constant is repaired. A stale root list is one possibility. Using f′ instead of f to generate a y-coordinate is another. The purpose is not to make the learner suspicious of everything. It is to separate consequences of the known correction from additional faults that require their own evidence.

If the learner cannot factor x² − 6x + 8, pause and teach or refresh that operation. Do not call the resulting delay a post-repair testing failure. If the learner factors correctly but accepts roots that violate a restriction, the next teaching action concerns domain filtering. A useful lesson distinguishes a missing mathematical tool from an inappropriate use of an available tool.

Minutes 20–35: model one complete recheck

Use one example with a visible change and a limited number of consequences. Work from the original question, not from the disputed line alone. State the corrected object, show the calculation that confirms it, identify the first downstream use, and explain why any preserved work is unaffected. Keep the explanation close to the mathematics.

For the constant-change example, the teacher might say: “Differentiating a constant gives zero, so this edit does not change the derivative. But evaluating the function includes the constant, so the old vertex height and tangent point need updating.” Then carry out the updates. This is a mathematical explanation of preservation, not permission to keep work because it looks familiar.

Include a test that is deliberately too weak. Checking f′(3) = 0 confirms stationarity at that x-value but cannot determine the vertex’s height. Ask the learner to repair the checking plan by adding evaluation in the original function. This makes the limitation of a passed check concrete and gives the student a reason to select another kind of evidence.

Minutes 35–55: rotate the solving and reviewing roles

For a pair, one learner repairs the supplied fault and the other names the next useful check before it is executed. For a group of three, use solver, reviewer and evidence recorder, rotating so each learner performs every role. The recorder’s task is narrow: note what the check establishes and whether any old value remains active after the edit.

The reviewer should not simply solve the entire question independently and announce a different answer. Ask for a targeted question: “Does this candidate meet the original denominator restriction?” or “Which equation produced this y-value?” Those prompts reveal whether the reviewer understands the repaired result’s consequences. A second full solution can be useful, but it is not the only possible review.

Use one item in which valid work should be preserved. For example, a corrected parameter may affect a tangent calculation but not a separately derived distance using unrelated coordinates. If every item requires restarting everything, the activity teaches excessive invalidation. Selective rechecking requires opportunities to keep correct work with a reason.

Also include one item where a formerly successful check is no longer current. A factorisation may have been verified before a coefficient was changed. The reviewer must notice that the tick belonged to the old polynomial. The lesson is about the relationship between evidence and the version tested, not about the physical presence of a tick on the page.

Minutes 55–70: remove the support

Give a different problem individually and remove role prompts and highlighted dependencies. Do not merely change the names in the previous worksheet. Change a feature that makes the learner select the check: an excluded denominator, an interval endpoint, a non-one-to-one substitution or a condition selecting a constant of integration.

For instance, after modelling a rational equation with an accepted candidate, use one whose reduced equation produces a forbidden value. The learner must return to the original domain rather than assume the earlier outcome will repeat. Keep the prerequisite algebra comparable so the new demand can be interpreted sensibly.

Collect the first independent attempt before giving feedback. The question is whether the learner initiates the necessary check, not whether the learner can agree with it after the teacher points to the denominator. Record the scope of the evidence: one fresh rational equation, completed without a domain prompt. Do not inflate that into mastery of every equation type.

Minutes 70–85: feedback followed by another attempt

Give feedback on the missing claim. “You proved that the listed root works, but you did not show that your list is complete” is more actionable than “check more”. Similarly, “The derivative is correct; the y-coordinate came from the wrong function” directs attention to the specific handoff instead of sending the student back through every calculus line.

Ask for a short correction and then a changed attempt. If the teacher supplies the answer to the first item, the second item should require the learner to make the decision independently. Merely copying the corrected script establishes that a corrected script has been produced. It does not establish that the student can select the same check on a new problem.

A learner who already selected the right check but made an arithmetic slip needs different feedback. Preserve the good decision, correct the execution and test it again. Do not turn every wrong endpoint into evidence that the whole checking framework was misunderstood. Separating decision quality from calculation accuracy makes the teaching response more proportionate.

Minutes 85–90: create a useful exit record

Ask for three short entries: the correction made, the check that established its safe use, and one situation where that check would be insufficient. For a radical equation, the last entry might explain why checking only the squared equation can accept an extraneous candidate. For a tangent, it might explain why passing through the point alone does not establish the correct slope.

The record should point to the next practice item. “Recheck trigonometric endpoint coverage on a fresh closed interval” is useful. “Be less careless” is not a practice design. Keep the note small enough that the learner can actually use it in the next session.

Alicia’s route: make the downstream use visible

In our fictional example, Alicia repairs the source correctly but leaves old results farther down the page. Give her a short linked problem and ask her to underline every place that consumes the changed value. Start with two uses, then add a third non-adjacent use. Include one independent calculation so underlining everything is not a successful strategy.

Her next goal is not a larger quantity of mathematics. It is a complete source-to-use update on one suitable problem. Ask her to state the current value before continuing and to verify the first important use after the edit. Gradually remove the underlining prompt. The independent performance being sought is identification of actual dependents, not beautiful annotation.

Tricia’s route: preserve work with a mathematical reason

Tricia’s fictional difficulty is repeated full restarting. Give her a correction with a proved invariant, such as a vertical constant change that leaves the derivative unchanged. Require her to name something that must change and something that must not change. She should then carry out only the necessary update and explain why the preserved result still applies.

Do not demand minimal editing at the expense of clarity. Sometimes rewriting a short subpart from a clean line is safer than maintaining several ambiguous versions. The aim is proportionate rechecking. A slightly wider clean restart can be justified; indiscriminately repeating every unaffected calculation is what the exercise is designed to question.

Kai Kai’s route: test the claim, not the familiar endpoint

Kai Kai’s fictional response is to trust an expected number even when the route is invalid. Use an example where two sign mistakes cancel, or where several listed trigonometric angles are valid but another required angle is missing. Hide the final answer while he explains the route and the limits of each check.

Then reveal a correct endpoint paired with invalid working. Ask which statement would fail on a new use of the intermediate formula. For a derivative with an unjustified final sign reversal, a second evaluation can reveal the surviving workaround. The exercise should distinguish genuine invariance from accidental compensation rather than treating every repeated sign change as a fault.

Adapt the format without changing the mathematical standard

A learner may explain a checking decision orally before recording it in writing. Another may need a larger, less crowded copy of the script to identify stale values. Use those adjustments to make the mathematical decision observable. Do not infer a conceptual defect merely from difficulty reading dense overwritten working.

At the same time, prepare for the actual permitted examination conditions. A practice conversation with a partner is not independent timed work. Record the support used and gradually move towards the required written performance. The standard remains the validity of the mathematical claim, while the teaching route can vary.

For a shorter session, select fewer problem families rather than compressing every stage into rushed instructions. A well-taught domain-sensitive example, followed by one independent contrast item, can provide a clearer next teaching decision than ten unrelated faults discussed superficially. This is a proposed prioritisation of lesson time, not a claim about a universally optimal number of exercises.

The lesson is ready to end when the teacher can identify what the learner did independently, what still required support and what fresh task will test the next step. It is not ready merely because every worksheet has been corrected in the teacher’s handwriting. The visible outcome should be a learner’s defensible checking decision attached to actual mathematics.

25. What one successful recheck does and does not show

A repaired solution and a repaired learning habit are different achievements. The first concerns one piece of mathematics: the current working now supports its conclusions. The second concerns a learner’s future behaviour: will they initiate an appropriate check on another problem without the same prompts? Evidence for the first should not be silently promoted into evidence for the second.

This distinction does not diminish a successful correction. Finding and fixing an error is worthwhile. It becomes even more useful when the learner can explain what changed and why the chosen check was sufficient. But one successful explanation immediately after a model answer is a narrower observation than independent use on a new task later.

Describe the observation at its actual scope

A precise record might say: “On this rational equation, the learner independently restored the denominator restriction after repairing the factorisation and rejected the forbidden candidate.” That statement names the task family, the decision and the level of support. It does not claim that the learner will now handle every radical, logarithmic and trigonometric equation correctly.

Compare it with “The learner has mastered checking.” The latter phrase hides too much. Did the learner choose the check, execute it, interpret the result, or merely follow an instruction? Were all candidates tested? Was the example already familiar? Was the final answer available? A useful record makes these conditions visible instead of compressing them into a global label.

When the purpose is to validate the current script, exact reasoning can be conclusive. A complete factorisation on the permitted domain, with valid candidate filtering, can establish the solution set for that equation. When the purpose is to predict future performance, one solved example is not a universal theorem about the person. The kind of claim determines the kind of evidence required.

Separate assisted correction from independent checking

Suppose a teacher says, “Remember that x cannot equal 2,” and the learner then rejects x = 2. The final answer may now be correct. What the observation shows is successful use of a supplied restriction. It does not show that the learner independently recognised the need to return to the original denominator.

A later task can test that initiation separately. Present a rational equation with a different excluded value and no domain reminder. Ask the learner to solve and check it. If the restriction is carried through unaided, record that additional evidence. There is no need to deny the value of the earlier assisted work; simply avoid counting it as a different kind of performance.

The same principle applies to collaborative practice. A group can produce a correct script while one learner relies on a partner’s choice of check. Rotate roles and collect an individual attempt. The group result and individual result answer different questions. Neither should be used to make unsupported claims about the other.

Change a feature that matters

A fresh problem should require a genuine decision, not just a change of names. After an equation where cancellation produces an allowed root, use one where it produces a forbidden root. After a half-open trigonometric interval, use a closed interval whose upper endpoint is actually a solution. After a positive transformed square value, include an impossible negative candidate.

These changes are purposeful because they challenge a simplistic rule. “Accept the result after cancelling” fails when the candidate is excluded. “Discard the repeated direction at 2π” fails when both closed endpoints are permitted inputs. “Take a positive square root” fails when a substitution has two real preimages and the original domain includes both.

Do not change every feature simultaneously at first. If the new task introduces unfamiliar algebra, a novel graph representation and a much shorter time limit, a failure is hard to interpret. Start by changing the checking demand while keeping prerequisite mathematics reasonably comparable. Then broaden the variation as the learner’s actual work justifies it.

A mathematically identical problem with different coefficients can still be useful for fluency. It simply provides narrower evidence of transfer than a contrast that requires a different acceptance decision. Match the follow-up task to the question you are trying to answer about the learner.

Use delayed and mixed opportunities without inventing an optimum

Return to the relevant checking decision in a later session, and eventually place it among problems that do not announce the topic. The IES study and instruction guide supports spacing learning and using retrieval opportunities in general. It does not prescribe one ideal interval for this particular learner or this particular correction routine.

A practical plan might include an independent contrast item in the next lesson and another relevant opportunity in a later mixed set. Treat those timings as adjustable planning choices. If the learner still cannot execute the underlying algebra, teach that prerequisite rather than interpreting a delayed failure as proof that the entire checking habit has disappeared.

A mixed task adds a useful demand: recognise that a check is needed without a heading such as “extraneous roots” giving the answer away. But it should remain within the learner’s taught content. Removing a topic label is not the same as introducing unannounced new mathematics.

Count opportunities before calculating a rate

During review, record both how often a relevant checking opportunity arose and how often the learner handled it independently. A count of four successful checks means something different if there were four opportunities, ten opportunities, or four repetitions of the same already corrected example.

For illustration only, imagine six fresh tasks each required a domain decision. The learner initiated the appropriate check independently in four, used a teacher prompt in one and missed it in one. Report those three categories. Do not record five independent successes by combining the prompted response with the four unaided responses.

Execution can be recorded separately. Perhaps all four independently chosen checks were appropriate, but one contained an arithmetic slip. That gives four sound checking decisions and three correctly executed independent checks in this small sample. The distinction points to a more precise next lesson than one undifferentiated percentage.

Do not interpret these illustrative counts as a statistical estimate of a learner’s future exam probability. The tasks may not be random, equally difficult or independent. A descriptive record is still useful, but the denominator and task conditions should travel with the result.

Measure efficiency only while preserving validity

A learner may reduce checking time by omitting completeness or domain filtering. That is not the improvement sought here. Compare the time taken to establish the required claims, not simply the time taken to put a tick beside the final number. A fast incomplete check and a slower complete check are not equivalent performances.

One practical review method records the selected check, whether it could detect the relevant fault, whether it was executed correctly and whether the learner stopped after sufficient evidence. Time can be added to that record. It should not replace the other entries.

Also record unnecessary repetition when it is visible. A learner who recomputes an independent distance three times after changing a logarithm’s domain may be using checking time without addressing the actual edit. The feedback should redirect that effort to the affected claim, not simply tell the learner to work faster.

A later wrong answer needs a fresh diagnosis

Suppose the learner handles a denominator restriction independently today but makes a later sign error in a derivative. The later error does not erase the earlier evidence. It is a different failure that may require a different teaching response. Conversely, another domain error may look similar but arise from a different stage, such as misunderstanding the original expression rather than forgetting a known restriction.

Inspect the current script. Identify the changed object, the test chosen and the precise failure. Do not automatically apply the last lesson’s diagnosis to every new wrong answer. The existing Error Signature Analysis guide can help organise hypotheses, but a pattern remains evidence to test rather than a guaranteed cause.

A correct answer needs the same care in interpretation. It might arise from a sound route, a familiar rehearsed procedure, a supplied hint or compensating errors. Inspecting the route allows a more accurate account of what succeeded. Do not punish correct work by assuming luck, but do not claim understanding solely from numerical agreement either.

Distinguish improvement after teaching from proof of causation

If a learner performs better after a lesson, it is reasonable to record the sequence: the lesson occurred, then these tasks were completed more successfully. That sequence alone does not establish how much improvement was caused by this particular routine. Other practice, task difficulty, teacher prompts and prior familiarity may have changed as well.

The practical response is to keep the claim bounded and collect useful comparison evidence. Use comparable fresh tasks, record support and inspect the actual checking decisions. This can guide teaching without pretending that an informal classroom comparison is a controlled study.

The article’s fictional scenes are not evidence of measured outcomes. Alicia, Tricia and Kai Kai make different decisions visible; they do not supply testimonials or data. Any real learner record should identify what was observed rather than borrowing the fictional characters’ improvement as an expectation.

Know when to close the current check

Closing a corrected script is a mathematical decision. The source has been restored, its affected uses have been checked, the original conditions are met and the answer has the required form. Once those claims are adequately established, the learner can continue. No additional feeling of absolute reassurance is required.

Closing a learning intervention is a broader educational decision. It requires evidence across appropriate independent opportunities and should be considered with the learner’s wider work. This chapter marks that boundary rather than creating a competing whole-course mastery system. The Reliability Testing guide addresses repeatability of examination-ready methods.

For teaching decisions about whether a repair survives a later complete paper, use the related Tutor Handbook discussion of the second full paper. That question is different from deciding whether the current tangent or candidate list has been properly rechecked after an edit.

The useful final record is specific: which checking decision is now independently demonstrated, which mathematical families have actually been tested, and what remains untested or still needs support. Precision makes progress easier to act on. It also protects the learner from both premature declarations of mastery and the discouraging claim that one later mistake means nothing was learned.

26. Teaching guide, evidence notes and final operating routine

A long reference guide should end with something a learner can actually do. This chapter provides one fresh exit task, an explained checking route and a compact record for deciding what to teach next. The aim is not to memorise the terminology in the chapter headings. It is to make a corrected result safe to use without losing valid work or accepting an unsupported conclusion.

The task below combines a parameter, a derivative, stationary points and a tangent. It deliberately excludes unfamiliar techniques so that the checking decisions remain visible. Select it only after these mathematical prerequisites have been taught. For learners not yet studying differentiation, use one of the algebraic contrast tasks in Chapter 23 instead; the same source-to-consequence reasoning can be practised there.

An independent exit task: one corrected parameter, several affected claims

Let f(x) = x³ − 3kx + 2 for real x. The original question states that f(2) = −2. Find k, the stationary points of the curve and their local classifications, and the equation of the tangent at x = 1.

A supplied draft originally used k = 3. Its author has now corrected the parameter to k = 2 but has left the old derivative 3x² − 9 and the old stationary x-values ±√3 on the page. A separate distance calculation between (0, 0) and (3, 4) is correctly recorded as 5. The author has not yet rechecked the tangent.

Before solving, decide what your next checks should be. Identify the original condition that confirms the parameter repair, the old calculations that depend on that parameter, and one result that can be preserved. Then produce the current mathematical answer. Do not look at the explanation until you have made those decisions.

Open the complete exit-task solution and retest plan

The defining condition is 8 − 6k + 2 = −2. Therefore 10 − 6k = −2, giving k = 2. Substituting it back gives f(2) = 8 − 12 + 2 = −2. This confirms the corrected parameter directly against the original condition, not merely against a later formula that used the old value.

The current function is f(x) = x³ − 6x + 2. Its derivative is f′(x) = 3x² − 6. The old derivative 3x² − 9 is correct for the old k = 3 function, but it is not the derivative of the current function. It must lose its role as the source for stationary points and tangent slopes.

Stationarity requires 3x² − 6 = 0, so x² = 2 and x = ±√2. Both signs are required because the original domain is all real x. The old values ±√3 fail the current stationary condition: substituting either into 3x² − 6 gives 3 rather than zero. This is a short test that directly exposes their stale status.

At x = √2, evaluate the original function, not its derivative. Since (√2)³ = 2√2, the y-value is 2√2 − 6√2 + 2 = 2 − 4√2. At x = −√2, the y-value is −2√2 + 6√2 + 2 = 2 + 4√2. The stationary points are therefore (−√2, 2 + 4√2) and (√2, 2 − 4√2).

The second derivative is f″(x) = 6x. It is negative at −√2 and positive at √2, so the first point is a local maximum and the second is a local minimum. A sign check on f′ gives the same local behaviour: positive outside the two stationary x-values and negative between them.

Do not describe these as a global maximum and minimum over all real x. The cubic is unbounded above and below. The question asks for local classifications, and the evidence establishes those classifications. The distinction protects a correct calculation from becoming an overstrong final claim.

For the tangent at x = 1, the current point is (1, −3), because f(1) = 1 − 6 + 2 = −3. The current slope is f′(1) = 3 − 6 = −3. Hence y + 3 = −3(x − 1), giving y = −3x.

The line passes through the required point and has the required slope. There is also an exact algebraic check: f(x) − (−3x) = x³ − 3x + 2 = (x − 1)²(x + 2). The repeated factor at x = 1 is consistent with equal function value and equal first derivative there. The additional intersection at x = −2 does not contradict tangency at x = 1.

The independent distance remains 5 because its two points do not use k, f or f′. Preserve it with that reason. A new parameter value does not invalidate every calculation written underneath it. Equally, the distance’s continued correctness supplies no evidence that the stationary points have been updated.

The current answer is k = 2; a local maximum at (−√2, 2 + 4√2); a local minimum at (√2, 2 − 4√2); and tangent y = −3x at x = 1. Parameter confirmation, derivative correction, original-function coordinates, local classification and tangent construction each have their own visible support.

Review the student’s decisions before grading the appearance of the page

Begin by checking whether the learner selected the right mathematical source. In the exit task, the condition f(2) = −2 determines k. A response that merely states “k = 2 was supplied in the correction” has not independently confirmed that repair. The student should be able to reconstruct the short defining calculation.

Next inspect the scope of the update. Did the learner regenerate the derivative and stationary values that used k? Did they evaluate the coordinates in f rather than in f′? Did they preserve the independent distance? These decisions are more informative than the number of crossed-out lines. A tidy page can contain stale mathematics, while a clearly corrected page can contain a sound argument.

Then inspect the interpretation. A correct stationary point can still be labelled with an unjustified global claim. A correct tangent can be rejected because it intersects the curve elsewhere. In both situations, the next teaching action concerns the test or interpretation being applied, not the arithmetic that already passed.

Finally, ask whether the learner can stop. Once the required parameter, points, classifications and tangent are established, repeated evaluation of the same distance has no new purpose. The learner should be able to point to the evidence for each required conclusion and return to the next task.

A response guide for the teacher

Observed response	What to inspect	A proportionate next teaching action
The parameter is corrected, but old stationary values remain	Whether the learner noticed the derivative’s dependence on k	Trace the parameter into the derivative and test one stale candidate
The x-values are correct, but the y-values are zero	Which function supplied the coordinates	Separate f(x) from f′(x) and evaluate the original function
The tangent has the correct slope through the wrong point	Whether both tangent inputs were updated	Check point membership and slope as separate conditions
The whole question is restarted, including the independent distance	Whether preservation has a mathematical justification	Identify one unaffected calculation and explain its independence
A valid tangent is rejected because it meets the curve elsewhere	The validity of the checking rule itself	Use the factorised curve-minus-line expression to clarify local tangency
The answer is correct only after a detailed prompt	The level of support used	Record guided success and select a fresh independent opportunity

This is an instructional response guide, not an official mark scheme or a validated psychological classification. Use it to select a next action on the observed mathematics. A learner who makes a new arithmetic slip while carrying out a sound checking plan should not automatically be treated as someone who failed to understand the plan.

A compact record that can travel to the next lesson

Use a small record with six fields: original task, material correction, affected claims, evidence collected, support used and next unresolved demand. Fill it with concrete mathematical objects. “Repaired k; refreshed derivative; checked both stationary coordinates; tangent point and slope pass” is more useful than “good correction lesson”.

A hypothetical exit record might read: “The learner independently confirmed k and regenerated f′, but used derivative values for the stationary points’ y-coordinates. After a prompt to use the original function, the coordinates were corrected. Next task: a fresh curve with stationary x-values already supplied, requiring independent selection of the function for y.” That record does not claim that a single prompted correction has established lasting independence.

For a learner who preserves too little, choose a contrast where a constant changes but the derivative does not. For a learner who preserves too much, choose one where a coefficient of x changes and the derivative must change. Pairing these cases teaches the mathematical reason for the decision instead of encouraging the rule “always redo the derivative” or its equally unreliable opposite.

Do not turn every field into another worksheet the learner must complete during the examination. The expanded record is for teaching and review. Exam use should compress into a current corrected line, the relevant checks and a clear answer. The purpose of preparation is to make the decision available, not to add paperwork when the learner needs to calculate.

Evidence notes: what supports this guide

The mathematical arguments and worked problems in this article are original instructional examples. They are explained so that the reader can inspect their validity directly. They are not copied examination items, official answer keys or reports of measured outcomes from real students. The recurring learners are fictional examples, not testimonials.

Cambridge IGCSE Mathematics – Additional (0606) provides a qualification-level reference for mathematical reasoning, analytical work and structured justification. Use the syllabus for the actual examination year when selecting examinable content. Its overview does not make every worldwide course’s assessment arrangements identical.

ISTQB Certified Tester Foundation Level syllabus, version 4.0.1, section 2.2.3, supplies the distinction between confirmation testing and regression testing that inspired the terminology. The application to handwritten Additional Mathematics solutions is an educational analogy. It is not an ISTQB-endorsed school programme or evidence of a specified improvement in examination marks.

OpenStax Algebra and Trigonometry 2e, “Other Types of Equations”, is a reference for radical-equation procedures and checking candidates in the original equation. The radical, rational and logarithmic examples here keep their own original domains explicit rather than treating a transformed equation as the entire problem.

OpenStax Calculus Volume 1, “Maxima and Minima”, provides the standard distinction between critical-point candidates and extrema. The article’s optimisation examples additionally establish feasibility and the relevant global conclusion, rather than treating every stationary point as the requested optimum.

The Institute of Education Sciences guide, Organizing Instruction and Study to Improve Student Learning, informs the broad lesson design. It does not validate this article’s precise session timing, fictional learner routes or named checking routine.

These sources serve different purposes. A qualification overview bounds content and assessment context. A mathematics text supports a mathematical convention or method. A teaching guide supports general instructional design. A software-testing syllabus supplies an analogy. None should be cited as though it proved every practical recommendation in the article.

Likewise, a school or tutor may adapt the exercises, but adaptation requires rechecking the mathematics. Changing a coefficient can introduce an excluded candidate, a repeated root or a new parameter regime. Changing an interval can add an endpoint. A teacher creating a new version should not keep the old answer key merely because the wording looks similar.

The distinction to remember: a valid answer and a complete answer

Many of the hardest checking failures in this guide arise from promoting one kind of success into another. A candidate satisfies an equation, so the learner assumes all candidates have been found. A line passes through a point, so the learner assumes it is the tangent. An antiderivative differentiates correctly, so the learner assumes its constant meets the given condition.

Each promotion needs an additional argument. Sometimes it is short: a nonzero linear coefficient gives uniqueness; two complete modulus branches establish coverage; a point and derivative slope determine the tangent. Sometimes it is longer, as in a parameter case distinction or a global optimisation argument. The amount of evidence follows the claim, not the learner’s wish to finish.

This is why rechecking should be selective but not superficial. Selectivity means omitting irrelevant repetition. It does not mean omitting the one domain or completeness argument that makes the conclusion valid. A five-second restriction check can be more important than several minutes repeating a factorisation that was already correct.

Continue through the Additional Mathematics performance library

Use the Additional Mathematics Hub for the broader learning route and Additional Mathematics Examination Performance for the overall relationship between knowledge and exam execution.

When several confirmed faults need correction, return to Repair Sequencing. When you cannot yet identify the fault, use Contradiction Localization. When a correct-looking answer may conceal invalid working, use Error Cancellation and Masking.

For the next layer of practice, connect Constraint Integrity with Case Coverage. The first protects the conditions under which a result may be used; the second protects the completeness of the cases and solutions that must be considered.

The final operating routine

Confirm the corrected source. Follow its actual uses. Reapply the original conditions. Establish the complete requested answer. Stop when those claims are supported; reopen the affected checks after another material edit.

Alicia no longer assumes that a new value at the top of the page has updated every later use. Tricia no longer discards a derivative merely because an unrelated constant changed. Kai Kai no longer treats one passing substitution as proof that an entire solution set is complete. Their fictional decisions point to three practical habits: update dependents, preserve justified invariants and keep every check within the scope of what it establishes.

A corrected line is the beginning of a repaired route, not always its end. The useful final question is simple: What evidence shows that the mathematics I am about to use belongs to the corrected problem, satisfies its conditions and supports the conclusion I am claiming? Answer that question with the working itself, and the correction is ready to carry forward.

Additional Mathematics Post-Repair Regression Testing | How to Recheck Corrected Working Before You Trust the Answer

The 50-second route

1. A corrected line creates a new checking question

2. Write a change record that remains useful

3. Find what the change can actually affect

4. Choose evidence that could catch the old mistake

5. Linear algebra and linked calculations

6. Rational equations and cancelled factors

7. Radicals, squaring and candidate validation

8. Logarithms and exponential substitutions

9. Modulus branches and inequality boundaries

10. Parameters that change the type of a question

11. Returning from a substituted variable

12. Trigonometric equations and complete angle sets

13. Multiple angles, division and lost solutions

14. Derivatives, stationary points and tangents

15. Integration, constants and signed area

16. Optimisation and feasible global answers

17. Coordinate geometry: equations must agree

18. Exactness, rounding and calculator handoffs

19. When correcting one error exposes another

20. Build a small but useful retest set

21. Use the method under examination time pressure

22. An integrated quadratic correction case

First correction: restore the constant

Second correction: restore the derivative

Rebuild and test the tangent

Rebuild and test the area

What the complete recheck establishes

23. Ten diagnostic exercises with explained answers

Exercise 1: the denominator that disappeared

Exercise 2: a valid root does not define every later expression

Exercise 3: a combined logarithm hides the original restrictions

Exercise 4: solving a branch is not accepting its candidate

Exercise 5: a parameter correction removes the equation’s variable

Exercise 6: the final variable and the stated domain both matter

Exercise 7: equal directions can be different permitted inputs

Exercise 8: a tangent can cross a curve

Exercise 9: three integration claims need three kinds of evidence

Exercise 10: a point check cannot certify a tangent’s direction

How to read your diagnostic result

24. A practical lesson and learner-specific routes

Before the lesson: select the mathematical demand

Minutes 0–8: collect an independent starting sample

Minutes 8–20: discuss the decision, not just the answer

Minutes 20–35: model one complete recheck

Minutes 35–55: rotate the solving and reviewing roles

Minutes 55–70: remove the support

Minutes 70–85: feedback followed by another attempt

Minutes 85–90: create a useful exit record

Alicia’s route: make the downstream use visible

Tricia’s route: preserve work with a mathematical reason

Kai Kai’s route: test the claim, not the familiar endpoint

Adapt the format without changing the mathematical standard

25. What one successful recheck does and does not show

Describe the observation at its actual scope

Separate assisted correction from independent checking

Change a feature that matters

Use delayed and mixed opportunities without inventing an optimum

Count opportunities before calculating a rate

Measure efficiency only while preserving validity

A later wrong answer needs a fresh diagnosis

Distinguish improvement after teaching from proof of causation

Know when to close the current check

26. Teaching guide, evidence notes and final operating routine

An independent exit task: one corrected parameter, several affected claims

Review the student’s decisions before grading the appearance of the page

A response guide for the teacher

A compact record that can travel to the next lesson

Evidence notes: what supports this guide

The distinction to remember: a valid answer and a complete answer

Continue through the Additional Mathematics performance library

The final operating routine

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