Solving the equation is not always the end of the problem. Mathematics can generate candidate answers that the original situation forbids: a zero denominator, a negative length, 7.4 buses, a probability above 1, a root outside the stated interval, or a time before the journey begins.
This fifty-fifth Secondary 4 Mathematics Learning Guide develops the filtering layer between algebraic solution and acceptable conclusion. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
It connects directly to Fractional Equations Reduced to Linear and Quadratic Equations, Linear Inequalities, Number Lines and Boundary Reasoning, and AO2 Model Building.
Candidate answer versus admissible answer
An algebraic method produces candidates. The original problem decides which candidates are admissible.
Calculate broadly. Accept narrowly.
The filtering criteria may come from the domain of a function, a denominator restriction, a physical condition, a counting requirement, a probability range, a stated interval or a geometric fact.
Denominators create forbidden values
Any value that makes an original denominator zero is outside the domain of the original expression, even if later algebra appears to produce it.
Worked Example 1 | Reject a zero-denominator candidate
Solve x/(x−2)=2/(x−2).
Multiplying both sides by x−2 gives x=2.
But at x=2, the original denominator x−2 is zero.
Therefore x=2 is inadmissible and the original equation has no solution.
Restrictions should be written before manipulation
If the equation contains 1/(x+5), write x≠−5 before clearing denominators. This makes the restriction visible even after the denominator disappears from later working.
Square roots and geometry create non-negative quantities
Ordinary lengths, areas, times and masses are non-negative. An algebraic root that violates the physical interpretation must be rejected.
Worked Example 2 | Negative length
A rectangle has sides x and x+3 and area 40 cm².
x(x+3)=40, so x²+3x−40=0.
(x+8)(x−5)=0.
Candidates: x=−8 or x=5.
A side length cannot be −8 cm in this model.
x=5 cm.
Counts create integer constraints
People, buses, chairs, boxes and teams usually require whole-number answers. A decimal may be mathematically correct before interpretation but not admissible as the final count.
Worked Example 3 | Minimum number of buses
287 students travel in buses carrying at most 40 students each.
287/40=7.175.
Seven buses carry at most 280 students, which is insufficient. Therefore:
8 buses are required.
This is not ordinary rounding to the nearest integer. The capacity constraint forces upward rounding.
Probability has a built-in domain
Every probability p must satisfy:
0≤p≤1.
If algebra produces p=1.3, the candidate is impossible as a probability. The error may come from selecting the wrong root, building the wrong equation, or interpreting a non-probability quantity as p.
Worked Example 4 | Filter a probability root
An equation for a probability gives p²−1.2p+0.2=0.
Factor:
(p−1)(p−0.2)=0.
Both p=1 and p=0.2 lie in [0,1], so both are mathematically admissible unless the original event conditions remove one.
The probability range is necessary, but sometimes not sufficient; the original context still matters.
Stated intervals can reject otherwise valid roots
A question may specify 0≤x≤10. If the equation has roots x=4 and x=13, only x=4 belongs to the required interval.
Worked Example 5 | Filter by interval
Solve x²−17x+52=0 for 0≤x≤10.
(x−4)(x−13)=0.
Candidates are 4 and 13. Only 4 satisfies the stated interval.
Boundary words determine whether endpoints are included
- “At least 5” → x≥5.
- “More than 5” → x>5.
- “At most 5” → x≤5.
- “Less than 5” → x<5.
Worked Example 6 | Capacity boundary
A lift carries at most 600 kg. Current load is 180 kg and each crate is 35 kg.
180+35n≤600.
35n≤420, so n≤12.
Because n is a non-negative integer, the maximum is 12 crates.
Boundary cases are powerful checks
A boundary case is an input at the edge of what the model allows. Testing it can expose whether the equation, inequality or interpretation behaves correctly.
Worked Example 7 | Test the exact boundary
For the lift model above, n=12 gives:
180+35(12)=600.
The boundary is included because the wording says “at most 600 kg”. If it said “less than 600 kg”, n=12 would be inadmissible.
Extraneous roots can arise from algebraic transformations
Operations such as squaring both sides can introduce solutions that satisfy the transformed equation but not the original equation.
Worked Example 8 | Check after squaring
Solve √(x+1)=x−1.
The right side must be non-negative, so x≥1.
Square both sides:
x+1=(x−1)²=x²−2x+1.
x²−3x=0, so x=0 or x=3.
x=0 violates x≥1 and also fails the original equation. x=3 works.
x=3.
Geometry can make two algebraic roots represent one physical configuration
Sometimes two roots do not represent two different objects. They may correspond to swapping labels.
Worked Example 9 | Same rectangle, swapped sides
A rectangle has perimeter 30 m and area 44 m². Let one side be x, so the other is 15−x.
x(15−x)=44 gives x=4 or x=11.
These are the same side pair, 4 m and 11 m, with the labels reversed. The two algebraic roots correspond to one physical rectangle.
Domain can come from the model, not only the formula
Suppose a linear growth model is stated only for 0≤t≤12. Evaluating it at t=100 is algebraically possible, but outside the model’s declared range. The number can be computed without being justified.
Worked Example 10 | Valid equation, invalid extrapolation
h=14+3t is used for a plant over the first 12 weeks.
At t=100, the equation gives h=314. But unless the model has been validated that far, this is an extrapolation outside the stated domain.
Computable does not automatically mean admissible.
Trigonometric answers also have geometric constraints
An angle in an ordinary triangle must lie between 0° and 180°. Side lengths must be positive and must satisfy triangle conditions.
Worked Example 11 | Triangle inequality check
Can sides 3 cm, 4 cm and 9 cm form a triangle?
No. The sum of the two shorter sides is 7 cm, which is less than 9 cm.
The proposed dimensions are geometrically impossible.
Financial answers need contextual admissibility too
A repayment period of 17.4 months may need interpretation if payments occur monthly. A required purchase quantity may need upward rounding if items are sold only in whole packs. A discount cannot reduce a positive price below zero under an ordinary percentage model.
Worked Example 12 | Whole packs
Each pack covers 2.5 m². A floor requires 23.1 m².
23.1/2.5=9.24 packs.
You cannot purchase 0.24 of a sealed pack under the stated model.
10 packs are required.
The admissibility audit
- What candidates did the mathematics produce?
- What restrictions existed before manipulation?
- Can any denominator be zero?
- Must the answer be positive or non-negative?
- Must it be a whole number?
- Does probability require 0≤p≤1?
- Is there a stated interval or model range?
- Are boundary endpoints included?
- Does the geometry physically exist?
- Does substitution back into the original problem confirm the candidate?
Common failure modes
| Failure | Cause | Repair |
|---|---|---|
| Accepts all quadratic roots | Context ignored after algebra | Filter through original conditions |
| Forgets denominator restrictions | Restriction disappears after clearing fractions | Write exclusions first |
| Rounds counts normally | Discrete constraint ignored | Use minimum/maximum logic |
| Accepts p>1 | Probability domain forgotten | Enforce 0≤p≤1 |
| Uses root outside stated interval | Local domain skipped | Check interval at the end |
| Evaluates model far outside range | Equation confused with validated reality | Respect model domain |
| Fails to substitute back after squaring | Extraneous root risk ignored | Verify in original equation |
Independent practice
- Solve x/(x−4)=4/(x−4) and state the admissible solution set.
- A rectangle has sides x and x+5 and area 24. Solve and interpret x.
- 253 people need vehicles carrying 32 people. Find the minimum number.
- An algebraic probability equation gives p=−0.1 or p=0.6. Which is admissible?
- Solve x²−14x+45=0 for 0≤x≤6.
- A lift limit is strictly less than 500 kg. Current load is 150 kg and each box is 35 kg. Find maximum boxes.
Explained answers
1. Algebra gives x=4, but x=4 makes the original denominator zero. Therefore no solution.
2. x(x+5)=24 → x²+5x−24=0=(x+8)(x−3). Candidates −8,3; length requires positive value, so x=3.
3. 253/32≈7.91, so 8 vehicles.
4. p=0.6.
5. (x−5)(x−9)=0. Only x=5 lies in 0≤x≤6.
6. 150+35n<500 → 35n<350 → n<10. Maximum integer=9.
Final thought
Mathematics produces possibilities; the original problem supplies permission. A complete solution therefore ends only after candidate answers have been tested against restrictions, domains, boundaries and the physical meaning of the situation.
The last step is not “I found x.” The last step is “This x is allowed.”
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