Most tested Papers 1 & 2

Algebra and Functions: Indices, Surds, Quadratics, Polynomials and Mappings

Algebra and functions underpins TMUA Section 1: indices, surds, quadratics, inequalities, polynomials and mappings. Because the test is no-calculator and multiple-choice, aim for exact values and fast reasoning via the discriminant, completing the square and the factor theorem.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • The index laws hold for all rational exponents: multiplying adds powers, dividing subtracts them, and a power of a power multiplies them.
    a^m × an = am+n
  • A negative index means take the reciprocal and a fractional index means take a root (a-1/2 = 1/√(a)).
  • Simplify surds by extracting square factors, and rationalise a denominator by multiplying top and bottom by the conjugate (√(12) = 2*√(3)).
  • The discriminant b2 - 4ac gives the number of real roots without solving: > 0 means two distinct roots, = 0 one repeated root, < 0 none.
  • Completing the square rewrites a quadratic as a(x - h)2 + k, exposing the turning point (h, k) and the minimum or maximum value.
  • Solve a linear-and-quadratic pair by substituting the linear equation into the quadratic to get a single equation in one unknown.
  • For a quadratic inequality, find the roots then use the parabola shape: an upward parabola is positive outside its roots and negative between them.
  • Factor and Remainder Theorems: the remainder when f(x) is divided by (x - a) is f(a), and (x - a) is a factor exactly when f(a) = 0.

Diagram

xy024-404y = (x-2)^2 - 5(2, -5)
Completing the square gives the vertex form: y = (x - 2)2 - 5 has its minimum at the vertex (2, -5).

Formulae

a^m × an = am+n, (a^m)n = amn, a-n = 1/an

Combining or simplifying powers that share the same base; a0 = 1 for any non-zero a.

am/n = (a1/n)^m = (a^m)1/n

Converting between a rational index and root or surd form, e.g. 82/3 = (cube root of 8)2 = 4.

x = (-b ± √(b2 - 4ac))/(2a)

Solving a*x2 + b*x + c = 0 when it does not factorise neatly; leave the surd exact.

a*x2 + b*x + c = a(x + b/(2a))2 + c - b2/(4a)

Completing the square to find the turning point and the minimum or maximum value of a quadratic.

1/(p + √(q)) = (p - √(q))/(p2 - q)

Rationalising a denominator by multiplying by the conjugate, using the difference of two squares.

b2 - 4ac

Deciding root count fast: > 0 two distinct real roots, = 0 one repeated root, < 0 no real roots.

Definitions

Surd
An irrational root left in exact form, such as √(2) or 3*√(5); kept exact rather than written as a rounded decimal so that no accuracy is lost.
Discriminant
The quantity b2 - 4ac taken from a quadratic a*x2 + b*x + c = 0; its sign alone tells you whether there are two, one, or no real roots.
One-to-one vs many-to-one mapping
A function is one-to-one if every output comes from exactly one input; it is many-to-one if some output value is produced by two or more inputs, as with x2 and |x|.
Factor Theorem
For a polynomial f(x), the linear expression (x - a) is a factor if and only if f(a) = 0; this lets you test candidate factors by substitution instead of dividing.

Worked examples

1

Find the set of values of k for which x2 + k*x + 9 = 0 has two distinct real roots.

  1. 1

    Two distinct roots need:

    b2 - 4ac > 0.

  2. 2

    Substitute a = 1, b = k, c = 9:

    k2 - 36 > 0.

  3. 3

    Factorise the left-hand side:

    (k - 6)(k + 6) > 0.

  4. 4

    This is an upward parabola in k, positive outside its roots -6 and 6.

  5. 5

    Read off the region:

    k < -6 or k > 6.

Answer: k < -6 or k > 6.

2

Write (3 + √(5))/(√(5) - 1) in the form a + b*√(5), where a and b are integers.

  1. 1

    Multiply top and bottom by the conjugate √(5) + 1.

  2. 2

    Expand numerator:

    (3 + √(5))(√(5) + 1) = 3*√(5) + 3 + 5 + √(5) = 8 + 4*√(5).

  3. 3

    Expand denominator:

    (√(5) - 1)(√(5) + 1) = 5 - 1 = 4.

  4. 4

    Simplify the fraction:

    (8 + 4*√(5))/4 = 2 + √(5).

Answer: 2 + √(5), so a = 2 and b = 1.

Common mistakes

  • ×Mixing up the discriminant conditions: two distinct roots need b2 - 4ac > 0, a repeated root needs = 0, and no real roots needs < 0.
  • ×Solving (k - 6)(k + 6) > 0 as if the answer were between the roots; an upward parabola is positive OUTSIDE the roots, so it is k < -6 or k > 6, not -6 < k < 6.
  • ×Treating √(x2) as x: it equals |x|, and a1/2 denotes only the non-negative root.
  • ×Losing the cross term when expanding: (a - b)2 = a2 - 2*a*b + b2, and the same 2*a*b appears when completing the square.
  • ×Misreading indices: a-n = 1/an (not -an), and (a^m)n = amn (not am+n).

No-calculator tips

  • ✓To answer 'how many roots', just test the sign of b2 - 4ac rather than solving the whole equation.
  • ✓Clear surds from a denominator with the conjugate and spot the difference of two squares so the bottom becomes a whole number.
  • ✓Before long division, use the Factor Theorem: test small values like ±1, ±2 and the factors of the constant term to find a root quickly.
  • ✓For a quadratic inequality, sketch the parabola: upward parabola positive outside the roots, negative between them; flip this for a downward one.
  • ✓Keep every answer exact as a surd or fraction; multiple-choice distractors are often the rounded value or a sign-flipped version of the correct one.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.Evaluate 27^(-2/3) × 16^(3/4).

  • A. 72
  • B. 8/9
  • C. -8/9
  • D. 1/72
Show answer

Answer: B — 8/9

27^(-2/3) = 1/(32) = 1/9 and 16^(3/4) = 23 = 8, so the product is 8/9. The 72 trap drops the reciprocal that the negative index requires; -8/9 wrongly treats a negative index as a negative sign.

Q2.For which values of k does x2 - k*x + (k + 3) = 0 have NO real roots?

  • A. k < -2 or k > 6
  • B. -2 ≤ k ≤ 6
  • C. -6 < k < 2
  • D. -2 < k < 6
Show answer

Answer: D — -2 < k < 6

No real roots needs discriminant < 0: k2 - 4(k + 3) = k2 - 4k - 12 < 0, i.e. (k - 6)(k + 2) < 0, giving -2 < k < 6. The ≤ version wrongly includes the equal-root (still real) cases, and the outside region is where there ARE two real roots.

Q3.When f(x) = 3x3 - 2x2 + c*x - 4 is divided by (x - 1), the remainder is 6. Find c.

  • A. 9
  • B. 3
  • C. -15
  • D. 5
Show answer

Answer: A — 9

By the remainder theorem f(1) = 6: 3 - 2 + c - 4 = c - 3 = 6, so c = 9. Setting the remainder to 0 gives the c = 3 trap, using x = -1 gives -15, and forgetting the -4 term gives 5.

Read this topic in the official UAT-UK TMUA content specification →

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