IB Mathematics: Applications and Interpretation
AI HL Prior Learning Formulae
A focused, exam-ready guide to the five higher-level entry skills the IB explicitly assumes: greatest common factors and least common multiples, rational exponents, rationalising denominators, quadratic equations and inequalities, and algebraic fractions.
Reviewed against the current IB Mathematics: Applications and Interpretation guide. This guide explains prerequisite knowledge and should be used with the syllabus and formula booklet for your own examination session.
Students often begin Mathematics: Applications and Interpretation Higher Level expecting statistics, modelling and technology to dominate every lesson. Those are important features of the course, but success still depends on reliable algebra. A model may be created with a calculator, yet you must decide whether its domain is sensible. A quadratic graph may be produced instantly, yet you must interpret its roots and intervals. A numerical answer may be accurate, yet an exact radical can reveal structure that a decimal hides. The five HL-only prior-learning topics are the bridge between basic school mathematics and this more demanding reasoning.
This page is deliberately narrower and deeper than our complete AI SL and HL prior-learning guide. That shared guide covers number systems, units, accuracy, arithmetic, functions, geometry, trigonometry, statistics, probability and rates. Here, every main section concentrates on material the official prior-learning table specifically marks as HL only. The goal is not to memorize isolated tricks. It is to understand the structure well enough to select a method, carry it out accurately, check it with technology and communicate the result.
What “AI HL-only prior learning” actually means
The IB guide explains that students arrive with different mathematical experiences, but examination questions assume familiarity with the listed prior-learning topics. Teachers are expected to identify missing knowledge and address it early. “HL only” in this context means the skill is an assumed prerequisite for Higher Level students; it does not mean the skill exists as an isolated assessed chapter or that every examination must include a direct question on it.
The official list identifies five additions for AI HL: greatest common factor or divisor and least common multiples; evaluation of exponential expressions with rational exponents; rationalising the denominator; solving quadratic equations and inequalities with rational coefficients; and adding or subtracting algebraic fractions. Simplifying simple roots is common prior learning, but the additional act of rationalising a denominator is specifically marked HL only. Likewise, all students meet linear equations, while the stronger quadratic equation and inequality requirement is marked HL only.
| HL-only prerequisite | Minimum evidence of readiness | Why it matters later |
|---|---|---|
| GCF and LCM | Use prime powers or Euclidean reasoning and interpret the result | Number structure, cycles, exact simplification and algorithms |
| Rational exponents | Move fluently between powers and radicals, with domain awareness | Models, inverse processes, calculus and equation solving |
| Rationalising denominators | Use a radical or conjugate multiplier without changing value | Exact forms, equivalent expressions and algebraic control |
| Quadratics | Solve equations and sign-based inequalities using several representations | Functions, optimization, intersections and model interpretation |
| Algebraic fractions | Factor, state restrictions, find an LCD and simplify safely | Rational functions, equations, rates and symbolic reasoning |
AI does not mean “algebra is optional.” Technology changes which calculations are efficient, but it raises the importance of decisions: what expression should be entered, which window should be used, which root belongs to the context, and whether a decimal is an approximation or an exact value. Weak symbolic work makes those decisions fragile. Strong symbolic work lets technology become evidence rather than authority.
A focused AI HL readiness diagnostic
Attempt the following without notes or a computer algebra system. A calculator may be used only after you have written a method. Label each item secure, slow or not yet secure. Correctness matters, but so do restrictions, interval notation and explanations.
- Write 756 as a product of prime powers.
- Find the GCF and LCM of 84 and 126.
- Evaluate 813/4 exactly.
- Simplify 322/5 and explain which operation you perform first.
- Simplify √72 and then rationalise 5/√8.
- Rationalise 3/(2+√5) using a conjugate.
- Solve 3x²−5x−2=0 exactly.
- Solve x²−5x+6≤0 and express the answer as an interval.
- Simplify 2/(x−1)−1/(x+1), stating every excluded value.
- Solve 1/x+1/(x+2)=3/4 and reject any invalid solution.
Prime factorisation, greatest common factors and least common multiples
The structure behind a positive integer
Every integer greater than 1 can be expressed uniquely as a product of prime powers, apart from the order of the factors. This fundamental theorem of arithmetic turns questions about divisibility into questions about exponents. Repeatedly divide by the smallest possible prime, or build a factor tree, until every leaf is prime. Then collect repeated factors as powers.
A quick check is to multiply the powers: 4×27×7=756. A factor tree is not finished if one of its leaves is composite. For example, stopping at 21 is incomplete because 21=3×7. The number 1 is neither prime nor part of the prime factorisation.
Greatest common factor or divisor
The greatest common factor, also called the greatest common divisor, is the largest positive integer dividing each given integer. In prime-power form, keep only primes shared by all numbers and choose the smallest exponent. The word “greatest” refers to the result, while the smaller exponent ensures the factor fits inside every number.
Worked example: GCF of 84 and 126
126=2×3²×7
GCF=2×3×7=42
Check: 84/42=2 and 126/42=3. No larger shared factor is possible because 84 contains only one factor of 3 and 126 contains only one factor of 2.
The Euclidean algorithm is an efficient alternative for large values. Replace the larger integer by the remainder after division and repeat until the remainder is zero. The last non-zero remainder is the GCF. For 126 and 84, 126=1×84+42 and 84=2×42+0, so the GCF is 42. This method also reveals why the divisor is shared.
Least common multiple
The least common multiple is the smallest positive integer divisible by every given integer. In prime-power form, include every prime that appears and select the largest exponent. The result must contain enough copies of each prime to reproduce all numbers.
For two positive integers, GCF and LCM satisfy a useful check. This relationship is not a substitute for understanding, but it quickly detects arithmetic mistakes.
Here, 42×252=84×126=10,584. When a contextual problem asks when repeating events coincide, LCM is usually appropriate. When it asks for the largest identical group size, tile size or packet size with nothing left over, GCF is usually appropriate. Keywords help, but the decisive question is whether you want a shared divisor or a shared multiple.
Application: synchronized cycles
Two sensors report every 18 seconds and 30 seconds. If they report together now, when will they next coincide?
LCM=2×3²×5=90 seconds
The first common positive reporting time is 90 seconds. A GCF of 6 would describe a shared subdivision, not the next coincidence.
Negative integers are handled through their positive magnitudes unless a definition says otherwise. The GCF is conventionally positive. Zero requires care: every positive integer divides zero, GCF(a,0)=|a|, and LCM involving zero is often defined as zero. In an examination context, state the convention if it matters.
Rational exponents: powers that represent roots
A rational exponent is an exponent written as a fraction m/n, where m and n are integers and n is not zero. It connects exponent laws to radicals. The denominator gives the root and the numerator gives the power. For positive real a, either operation may be performed first.
Choose the order that keeps arithmetic simple. To evaluate 813/4, take the fourth root first: the fourth root of 81 is 3, and 3³=27. Raising 81 to the third power before taking a fourth root is valid but inefficient. For 322/5, the fifth root is 2, so the result is 2²=4.
Worked example: negative rational exponent
Evaluate 64−2/3.
The negative exponent creates a reciprocal. It does not make the final value negative.
Exponent laws still apply
For expressions within their real domains, rational exponents follow the usual index laws. Multiply like bases by adding exponents, divide by subtracting exponents, and multiply exponents when raising a power to a power. A zero exponent gives 1 when the base is non-zero.
a^p/a^q=a^(p−q), a≠0
(a^p)^q=a^(pq)
a^0=1, a≠0
Do not apply laws across addition. In general, (a+b)1/2 is not a1/2+b1/2. The square root of 9+16 is 5, while 3+4 is 7. Exponent laws describe products, quotients and nested powers, not sums.
Real-domain cautions
The denominator of the exponent affects which negative bases produce real values. An odd root of a negative number is real, but an even root is not real. Thus (−8)1/3=−2, while (−8)1/2 is not a real number. Parentheses matter: −82/3 may be interpreted as the negative of 82/3, whereas (−8)2/3 applies the power to the negative base.
Solving equations with rational exponents
Isolate the powered expression before applying an inverse power, and then check for extraneous or missing solutions. When both sides are known to be non-negative, operations are more predictable. Context may also restrict values, such as length, time or population being non-negative.
Worked example: isolate and invert
Solve x3/2=64 for real x where x≥0.
Check: 163/2=(√16)³=64.
Rational exponent fluency supports exponential models, power functions, derivative rules and transformations. It also helps you recognize that a radical expression and a fractional-power expression may describe the same function while displaying domain issues differently.
Simplifying surds and preserving exact values
A surd, or radical, is an exact expression involving a root that cannot be simplified to a rational number. The decimal approximation of √2 is useful for measurement, but √2 is the exact value. Exact forms prevent accumulated rounding error and expose algebraic relationships.
To simplify a square root, factor the radicand into a perfect square times another factor. Then take the square root of the perfect-square part. The same strategy works for higher roots using perfect powers.
cube root of 54=cube root of (27×2)=3 cube root of 2
Only like radicals combine under addition or subtraction. After simplification, 3√8+√18 becomes 3(2√2)+3√2=9√2. Before simplification the terms may look unlike, so always reduce their radicands first.
Worked example: expand and simplify
Simplify (√5+2)(√5−3).
Use ordinary expansion. Radicals behave like algebraic quantities, and √5×√5=5.
Radical rules and their conditions
For non-negative real a and b, √a×√b=√(ab). For b>0, √a/√b=√(a/b). These familiar rules require domain care. A frequent error is assuming √(a+b)=√a+√b, which is false. Another is writing √(x²)=x; over the real numbers the correct expression is |x| because the principal square root is non-negative.
√(a+b) is generally not equal to √a+√b
When a question requests an exact value, keep the radical. When a context requests a measurement, an appropriate decimal and unit may be needed. Good solutions often show both: an exact expression followed by a rounded approximation with a clear equality or approximation symbol.
Rationalising a denominator
Rationalising rewrites a fraction so that no irrational radical remains in its denominator. The value does not change because numerator and denominator are multiplied by the same non-zero expression. This creates an equivalent exact form that may be easier to compare or combine.
A single radical denominator
If the denominator is a single square-root term, multiply by that root over itself. Simplify the radical first if possible, then cancel common factors only after the algebra is valid.
Worked example: rationalise 5/√8
The final denominator is rational. A decimal check confirms both forms are approximately 1.7678.
Binomial denominators and conjugates
When the denominator has two terms and one contains a square root, multiply by the conjugate: keep the terms but reverse the sign between them. The middle radical terms cancel because the product is a difference of squares.
Conjugate of 2+√5 is 2−√5
Worked example: conjugate method
Rationalise 3/(2+√5).
=3(2−√5)/(4−5)=3√5−6
The denominator is −1, so the sign changes. A numerical comparison is an excellent check because the original denominator is positive and the result 3√5−6 is also positive.
Do not multiply only the denominator. Doing so changes the value. Do not use the same binomial instead of the conjugate, because squaring a binomial usually leaves a radical middle term. Finally, do not turn an exact answer into a decimal unless the question or context requires it.
Quadratic equations with rational coefficients
A quadratic equation can be written ax²+bx+c=0 with a≠0. Rational coefficients include integers and fractions. Before choosing a method, move every term to one side and simplify. Four main approaches are factorisation, completing the square, the quadratic formula and technology. A strong AI HL student can connect them rather than treating them as unrelated procedures.
Factoring and the zero-product property
If a product is zero, at least one factor is zero. Factorisation is efficient when integer or simple rational factors are visible. For 3x²−5x−2=0, find numbers whose product is −6 and sum is −5, then split the middle term.
x=−1/3 or x=2
Setting factors equal to zero is valid only when the equation equals zero. From (x−2)(x+4)=7, you cannot claim x=2 or x=−4. Expand or rearrange first.
Completing the square
Completing the square rewrites a quadratic in vertex form, which reveals the turning point and supports graph interpretation. For a monic quadratic x²+bx+c, add and subtract (b/2)².
Worked example: vertex form and roots
Rewrite x²−6x+5 and solve x²−6x+5=0.
(x−3)²=4 ⇒ x=1 or x=5
The vertex is (3,−4), and the symmetry of the roots around x=3 is visible.
The quadratic formula and discriminant
The quadratic formula solves every quadratic over the complex numbers and every real-solvable quadratic over the reals. Identify a, b and c carefully, including their signs. Use parentheses when substituting a negative coefficient.
The discriminant D=b²−4ac predicts the number of distinct real roots. If D>0 there are two; if D=0 there is one repeated root; if D<0 there are no real roots. In a graph, these cases correspond to two x-intercepts, one tangent intercept, or no x-intercepts.
| Discriminant | Real roots | Graph meaning |
|---|---|---|
| D>0 | Two distinct roots | Parabola crosses the x-axis twice |
| D=0 | One repeated root | Parabola touches the x-axis at its vertex |
| D<0 | No real roots | Parabola does not meet the x-axis |
When coefficients are fractions, you may multiply the entire equation by a common non-zero denominator to obtain integer coefficients. This does not change the solutions. Never clear only selected terms; every term must be multiplied.
Worked example: rational coefficients
Solve (1/2)x²−(5/6)x−1/3=0.
(3x+1)(x−2)=0
x=−1/3 or x=2
Technology can calculate or approximate roots, but a complete solution still records the equation, relevant roots and interpretation. If a model uses time, a negative root may be mathematically valid yet contextually impossible. Reject it with a reason rather than silently deleting it.
Quadratic inequalities and interval reasoning
A quadratic inequality asks where a quadratic expression is positive, negative, non-negative or non-positive. Solving the corresponding equation gives boundary points, but the solution is an interval or union of intervals. You must determine the sign between and outside the roots.
Worked example: upward-opening quadratic
Solve x²−5x+6≤0.
Roots: 2 and 3
Solution: 2≤x≤3
The parabola opens upward, so it lies on or below the x-axis between the roots. Equality includes both endpoints.
A sign chart provides an algebraic check. Test one value in each interval (−∞,2), (2,3) and (3,∞), or inspect factor signs. The product is positive outside the roots and negative between them. With a strict inequality, endpoints are excluded.
If the leading coefficient is negative, the parabola opens downward and the sign pattern reverses. If the discriminant is negative, there are no boundary roots, so the expression keeps the sign of the leading coefficient for all real x. If the discriminant is zero, the expression touches zero once and otherwise retains one sign.
Worked example: no real boundary points
Solve x²+4x+8>0.
Therefore x²+4x+8>0 for every real x.
Completing the square confirms this: (x+2)²+4 is always positive.
Graphing technology is useful for confirming intervals, but the viewing window must contain the relevant roots and turning point. A misleading window can make a crossing invisible. Solve or estimate key features first, set an appropriate window, and then use the graph as corroboration.
In contextual modelling, the solution must be intersected with the domain. Suppose an inequality is true for t<−1 or t>6, but t represents time after an experiment begins. The contextual solution is t>6, possibly with an upper limit determined by the model. Domain restrictions are part of the mathematics, not an optional sentence.
Adding and subtracting algebraic fractions
An algebraic fraction is a quotient of algebraic expressions. It is undefined wherever its original denominator equals zero. This restriction must be identified before cancellation because a simplified expression may hide a hole in the original domain. The workflow is consistent: factor denominators, state restrictions, find the lowest common denominator, rewrite each fraction, combine numerators, factor again and cancel only common factors.
Restrictions come from the original expression
For 1/(x−1)+2/(x+3), the excluded values are x=1 and x=−3. Even if later algebra cancels a factor, those values remain excluded from the original expression. A factor may cancel; a term cannot. You may cancel (x−2) in a product divided by (x−2), but not the x in (x+2)/x.
Different linear denominators
Worked example: subtract two fractions
Simplify 2/(x−1)−1/(x+1).
[2(x+1)−(x−1)]/[(x−1)(x+1)]
=(x+3)/(x²−1)
The subtraction sign applies to the whole second numerator. Writing 2x+2−x−1 would give the wrong constant; it must be 2x+2−(x−1)=x+3.
Factor before choosing a denominator
Expanded denominators can conceal shared factors. For example, x²−1=(x−1)(x+1). The lowest common denominator of 1/(x²−1) and 1/(x−1) is x²−1, not (x²−1)(x−1). Factoring prevents unnecessary expansion and reduces opportunities for error.
Worked example: shared factor
Simplify 1/(x²−1)+2/(x−1).
[1+2(x+1)]/[(x−1)(x+1)]
=(2x+3)/(x²−1)
Solving equations containing algebraic fractions
State restrictions first, then multiply every term by the lowest common denominator. Solve the resulting equation and check every candidate against the original expression. Multiplying by a variable expression can introduce a candidate at which the original equation was undefined.
Worked example: rational equation
Solve 1/x+1/(x+2)=3/4.
Multiply by 4x(x+2): 4(x+2)+4x=3x(x+2)
3x²−2x−8=0
(3x+4)(x−2)=0
x=−4/3 or x=2
Both values satisfy the restrictions and can be checked in the original equation.
Complex fractions
A complex fraction contains fractions in its numerator, denominator or both. Multiply the entire numerator and entire denominator by the internal lowest common denominator. Parentheses are essential because you are multiplying complete grouped expressions.
The original restrictions include x≠0, y≠0 and x≠y because the overall denominator must also be non-zero. Domain analysis is often the difference between a polished solution and an incomplete one.
These techniques lead directly into rational functions and algebraic manipulation used throughout AI HL Number and Algebra. They also appear when simplifying rates of change, rearranging models and comparing expressions produced by technology.
Technology, exact work and mathematical communication
AI HL expects purposeful technology use, but prerequisite questions still demand mathematical judgment. Use a three-layer workflow: obtain a symbolic or conceptual plan, use technology to compute or visualize, and independently validate the outcome. The order may change in an investigation, yet no single layer should be trusted blindly.
Before pressing Enter
Write the intended expression, add parentheses, identify the domain and predict the sign or scale. For a quadratic, estimate the number and location of roots from the discriminant or graph shape.
After the output
Substitute values back, compare exact and decimal forms, inspect residuals and explain any rejected root. A plausible screen value is not evidence until it is connected to the original problem.
Exact versus approximate answers
An equals sign connects equal quantities; an approximation sign marks rounding. If a root is x=(5+√13)/6, then x≈1.434 is an approximation. Keep guard digits until the final step and state the requested accuracy. In modelling, include units and interpret what the number represents.
When a calculator returns 1.9999999998 for an exact root of 2, numerical tolerance is involved. When it returns only one root because the graph window misses another, the issue is representation. When it produces an error for a rational power of a negative number, the input form or real-domain convention may be responsible. Each situation calls for mathematical diagnosis rather than repeated button pressing.
Graphing checks for quadratics
A reliable graphing check includes the vertex, intercepts and domain of interest. If the graph models profit only for 0≤x≤500, a root at x=−20 has no practical meaning even though it is algebraically correct. Trace or solve features with adequate precision, then communicate the contextual decision.
Continue from these foundations into the AI SL and HL Functions guide and the AI HL-only Functions guide. The prerequisite techniques on this page support those assessed function and modelling ideas; they do not replace them.
Integrated worked problems
Problem 1: common cycles and a quadratic model
A data logger samples temperature every 24 seconds and humidity every 36 seconds. Both sample at t=0. The simplified signal model is S(t)=−(1/144)(t−72)²+36 for 0≤t≤144. Find the first positive time at which both loggers sample together, then evaluate the signal.
LCM=2³×3²=72 seconds
S(72)=−(1/144)(0)²+36=36
The LCM finds the synchronization time. Substitution into vertex form shows the signal is at its maximum. Technology could verify the graph, but prime-factor reasoning and the model structure explain the result.
Problem 2: radical boundary
A threshold equation is 2x²−8=0. Solve exactly, write the positive root with a rational denominator if necessary, and give a decimal check.
This example simplifies without a surviving radical. If instead 3x²−2=0, then x=±√(2/3)=±√6/3. The rationalised exact form and decimal approximation ±0.8165 describe the same boundary. In a length context, only the positive root would remain.
Problem 3: rational equation becomes quadratic
Solve 2/(x−1)+3/(x+2)=2. State restrictions and verify the solutions.
2(x+2)+3(x−1)=2(x−1)(x+2)
5x+1=2x²+2x−4
2x²−3x−5=0
(2x−5)(x+1)=0
x=5/2 or x=−1
Both values respect the restrictions. Substitution confirms each. The problem joins algebraic fractions, a rational equation and quadratic factorisation—exactly the kind of dependency that makes prior learning important.
Problem 4: inequality with a contextual domain
A model is acceptable when E(t)=t²−9t+20≤0, where t measures days and 0≤t≤10. Factor and interpret.
E(t)≤0 for 4≤t≤5
The mathematical solution already lies inside the contextual domain, so the model meets the condition from day 4 through day 5, including both boundaries. A graphing check should show an upward parabola below the axis only between its roots.
Common mistakes and precise corrections
| Mistake | Why it fails | Better habit |
|---|---|---|
| Using the largest exponents for GCF | The result may not divide every number | GCF uses minimum shared exponents; LCM uses maximum exponents |
| Treating a negative exponent as a negative value | A negative exponent means reciprocal | Rewrite a−p as 1/ap |
| Distributing a root over addition | Radical product rules do not apply to sums | Test a simple numerical counterexample |
| Multiplying only the denominator while rationalising | The fraction's value changes | Multiply by an expression over itself |
| Finding roots but not intervals for an inequality | Roots are boundaries, not the full solution | Use a graph or sign chart and write interval conditions |
| Cancelling terms across addition | Cancellation applies to factors | Factor the entire numerator or denominator first |
| Forgetting original restrictions | A simplified form can conceal undefined values | State exclusions before manipulation and check candidates |
| Reporting calculator output without interpretation | The output may be approximate, extraneous or outside the domain | Substitute, check units and explain contextual decisions |
A fourteen-day AI HL prerequisite repair plan
This plan assumes 35 to 50 focused minutes daily. Start each session with five minutes of retrieval from the previous day and finish with one mixed problem. Record errors by cause—concept, algebra, notation, calculator input or interpretation—so revision addresses the real weakness.
- Day 1: complete the diagnostic and create an error log.
- Day 2: review primes, factor trees and prime-power notation.
- Day 3: practise GCF, LCM, Euclidean checks and contextual selection.
- Day 4: connect rational exponents to roots and apply index laws.
- Day 5: practise negative powers, rational-power equations and domain checks.
- Day 6: simplify radicals, combine like surds and preserve exact values.
- Day 7: rationalise single-term and conjugate denominators.
- Day 8: solve quadratics by factorisation and verify roots.
- Day 9: complete the square and connect vertex form to graphs.
- Day 10: use the quadratic formula and interpret the discriminant.
- Day 11: solve quadratic inequalities using sign charts and graphs.
- Day 12: factor denominators and add or subtract algebraic fractions.
- Day 13: solve rational equations, keeping restrictions visible.
- Day 14: complete a mixed set, use technology checks and redo every earlier error.
Frequently asked questions
Are these five topics the entire AI HL prior-learning list?
They are the topics specifically added or marked HL only in the official prior-learning table. AI HL students also need all shared prerequisites in number, algebra, functions, geometry, trigonometry, statistics, probability and rates. Use the linked shared guide for that broader checklist.
Will the examination ask a stand-alone GCF or LCM question?
It can assume the skill, but prior learning is not a promise of a dedicated question. GCF or LCM may appear inside a larger numerical, algorithmic or contextual problem. Learn to recognize the structure rather than waiting for a keyword.
Do I need to show exact answers if a calculator is allowed?
Follow the question. Exact forms remain important because they preserve structure and avoid premature rounding. When a contextual decimal is required, keep guard digits and round only at the end. Show whether a value is exact or approximate.
Why rationalise a denominator when a decimal is easier?
Rationalisation demonstrates equivalence and produces a conventional exact form. It also strengthens conjugate and difference-of-squares reasoning used elsewhere. A decimal may be useful for checking, but it does not replace the requested exact manipulation.
When should I factor a quadratic instead of using the formula?
Factor when factors are visible and efficient. Use completing the square when vertex information matters, and use the formula when factors are not obvious or an exact general method is needed. Technology can check any method.
How do I know whether a quadratic inequality is inside or outside the roots?
Inspect the leading coefficient and use a sign chart or graph. An upward-opening quadratic with two roots is negative between them and positive outside. A downward-opening quadratic reverses that pattern. Include roots only when equality is allowed.
Do cancelled denominator values become allowed again?
No. Restrictions come from the original expression. Cancellation may create a simpler rule, but it does not fill a hole that existed in the original domain. State excluded values before simplifying.
Can a calculator solve every rational-power equation correctly?
A calculator can help, but real-domain conventions, negative bases and input grouping can affect results. Write parentheses carefully, identify the domain and substitute solutions back. Mathematical control is essential.
Is simplifying radicals also HL only?
The current prior-learning table lists simplification of simple roots as shared knowledge. Rationalising the denominator is specifically marked HL only. HL students should be fluent with both because rationalisation depends on radical simplification.
What should I revise after this page?
Move to the dedicated AI Number and Algebra and Functions guides linked above, then practise mixed questions where algebra supports modelling, statistics or calculus. Continue using an error log so prerequisite slips do not hide inside advanced work.
Does the 2027 curriculum update remove the need for these foundations?
No sensible preparation would treat core algebra as disposable. The IB describes targeted reductions and first assessment in May 2029. Students should use the guide for their own session, but exact manipulation, domains and technology checks remain broadly valuable.
Is this an official IB publication?
No. This is an independent RevisionTown teaching guide informed by official curriculum documents. The International Baccalaureate Organization is the authoritative source for curriculum and assessment requirements.
Final AI HL prior-learning checklist
- I can express integers as prime powers and distinguish GCF from LCM contexts.
- I can evaluate positive, negative and fractional exponents with domain awareness.
- I can simplify radicals, combine like surds and preserve exact values.
- I can rationalise a single radical denominator and a binomial denominator using a conjugate.
- I can solve quadratics by factorisation, completing the square and the quadratic formula.
- I can interpret the discriminant and connect equations to parabola features.
- I can solve quadratic inequalities as intervals and apply contextual domains.
- I can add and subtract algebraic fractions, state restrictions and check rational-equation solutions.
- I can use technology to verify rather than replace reasoning.
- I can communicate exact versus approximate results, units and rejected values clearly.
Once these skills are secure, AI HL problems become more transparent. You can spend less time repairing algebra and more time modelling, interpreting and evaluating. Return to the shared AI prior-learning guide for the complete foundation, then build into the assessed formula guides as your course progresses.
Official curriculum references
- IB Mathematics: Applications and Interpretation guide
- IB Applications and Interpretation curriculum update
- IB Diploma Programme mathematics overview
RevisionTown is independent and is not affiliated with or endorsed by the International Baccalaureate Organization. Verify requirements against the documents for your examination session.
