The identities everything else follows from
- ✓(a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².
- ✓a² − b² = (a + b)(a − b).
- ✓(a + b)³ = a³ + 3a²b + 3ab² + b³ and (a − b)³ = a³ − 3a²b + 3ab² − b³.
- ✓a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).
- ✓(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
- ✓a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
- ✓Two consequences worth memorising: a² + b² = (a + b)² − 2ab, and (a + b)² − (a − b)² = 4ab.
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Linear equations and systems
Solve: 2x + 3y = 12 and x − y = 1.
x = 3, y = 2. From the second equation x = y + 1; substituting gives 2(y+1) + 3y = 12, so 5y = 10 and y = 2, hence x = 3. Check both equations: 6 + 6 = 12 ✓ and 3 − 2 = 1 ✓. Always substitute back - it costs seconds and catches sign errors.
The sum of two numbers is 25 and their difference is 7. Find the numbers.
16 and 9. For sum s and difference d, the numbers are (s + d)/2 and (s − d)/2 - here 32/2 and 18/2. This shortcut avoids setting up simultaneous equations at all.
For what value of k does the system 2x + 3y = 7 and 4x + ky = 14 have infinitely many solutions?
k = 6. Infinitely many solutions require the equations to be proportional: a₁/a₂ = b₁/b₂ = c₁/c₂. Here 2/4 = 7/14 = 1/2, so 3/k must equal 1/2, giving k = 6. If only the first two ratios matched, the system would be inconsistent instead.
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Quadratic equations
For a quadratic ax² + bx + c = 0: the discriminant is D = b² − 4ac, the sum of roots is −b/a and the product is c/a. A surprising share of questions never require you to find the roots at all.
Solve x² − 5x + 6 = 0.
x = 2 and x = 3. Factorise by finding two numbers with sum 5 and product 6. Always try factorisation before the quadratic formula - school-level quadratics are usually built to factorise.
Find k so that x² + kx + 9 = 0 has equal roots.
k = ±6. Equal roots require D = 0, so k² − 36 = 0. Note that both signs are valid answers, and dropping the negative one is a standard mark loss.
If α and β are the roots of x² − 7x + 12 = 0, find α² + β².
25. Use α + β = 7 and αβ = 12, then α² + β² = (α + β)² − 2αβ = 49 − 24. Finding the roots (3 and 4) also works, but the identity method generalises to cases where the roots are irrational.
If x + 1/x = 3, find x² + 1/x².
7. Square both sides: x² + 2 + 1/x² = 9, so x² + 1/x² = 7. Any symmetric expression in x and 1/x comes from squaring or cubing the given relation - never from solving for x.
Form the quadratic equation whose roots are 2 and −5.
x² + 3x − 10 = 0. Use x² − (sum)x + product = 0, with sum = −3 and product = −10. Watch the sign of the middle term: it is minus the sum.
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Related reading
Polynomials, factorisation and the remainder theorem
Factorise x² − 9x + 20.
(x − 4)(x − 5). Two numbers with product 20 and sum 9, both negative since the middle term is negative and the constant positive.
Find the remainder when x³ − 3x² + 4 is divided by (x − 2).
0. By the remainder theorem the remainder is f(2) = 8 − 12 + 4 = 0, so (x − 2) is a factor - which is the factor theorem as a special case. Long division is never needed for a linear divisor.
Factorise a³ + b³ + c³ − 3abc when a + b + c = 0.
It equals 0. Since a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), the first factor vanishes. Hence a + b + c = 0 implies a³ + b³ + c³ = 3abc - a result questions use directly.
Simplify (a + b)² − (a − b)².
4ab. Expanding both is unnecessary - recognise it as a difference of squares of the form X² − Y² = (X + Y)(X − Y) with X = a + b and Y = a − b, giving (2a)(2b).
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Sequences and series
Find the 15th term of the AP 3, 7, 11, ...
59. With a = 3 and d = 4, aₙ = a + (n − 1)d = 3 + 14 × 4. Note the (n − 1), not n - the single most common slip in this chapter.
Find the sum of the first 20 natural numbers.
210, using Σn = n(n+1)/2 = 20 × 21/2. Worth remembering alongside Σn² = n(n+1)(2n+1)/6 and Σn³ = [n(n+1)/2]².
Find the sum of the infinite GP 1 + 1/2 + 1/4 + 1/8 + ...
2, from S∞ = a/(1 − r) = 1/(1 − 1/2). The formula applies only when |r| < 1; for |r| ≥ 1 the series diverges and has no finite sum.
How many terms of the AP 5, 9, 13, ... add up to 165?
9 terms. Using Sₙ = (n/2)[2a + (n−1)d]: 165 = (n/2)[10 + 4(n−1)], so 330 = n(4n + 6), giving 4n² + 6n − 330 = 0, or 2n² + 3n − 165 = 0, which factors as (2n + 15)(n − 9) = 0. Only n = 9 is admissible, since a term count cannot be negative - always discard the inadmissible root explicitly.
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Inequalities, modulus and logarithms
Solve |2x − 3| = 7.
x = 5 or x = −2. A modulus equation splits into two cases: 2x − 3 = 7 and 2x − 3 = −7. Forgetting the second case loses half the answer.
Solve x² − 4 > 0.
x < −2 or x > 2. Factorise as (x − 2)(x + 2) > 0; the product is positive when both factors share a sign. Never 'take the square root' of an inequality - that loses the negative branch entirely.
Solve |x − 1| < 3.
−2 < x < 4. The rule |X| < a means −a < X < a, so −3 < x − 1 < 3. Note that 'less than' gives one interval while 'greater than' gives two - the asymmetry examiners test.
Find the value of log₂ 32 + log₃ 27.
8. log₂ 32 = 5 since 2⁵ = 32, and log₃ 27 = 3 since 3³ = 27. Log questions at this level are almost always about recognising a power.
If log x = 2 log 3 + 3 log 2, find x.
x = 72. Using n log a = log aⁿ, the right side is log 9 + log 8 = log 72, so x = 72. The three log rules - product, quotient and power - handle nearly every question in the chapter.
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How to practise algebra so it gets faster
- ✓Learn the identities in both directions. Recognising 4ab as (a+b)² − (a−b)² is what makes shortcuts available.
- ✓For any quadratic question, check whether sum and product of roots suffice before solving for the roots.
- ✓Substitute your answer back into the original equation, every time. It catches almost every sign error.
- ✓Discard inadmissible roots explicitly - negative counts, negative lengths, values that make a denominator zero.
- ✓Never square or square-root an inequality. Factorise and reason about signs instead.
- ✓Practise timed once the methods are secure, since algebra underpins the speed of every other chapter.
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