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O-Level Additional Mathematics · Secondary 3 and 4

Surds

A surd is a root that cannot be simplified to a rational number, such as √2, √5 or √12. In A-Math you simplify surds, rationalise denominators and solve equations that contain surds, all without a calculator. Surd skills reappear in trigonometry, where exact values such as sin 60° = √3/2 come up, in coordinate geometry when you find exact lengths, and in quadratic equations whose roots are not whole numbers. Questions often ask for an answer in the form a + b√c, so exact working matters.

What the syllabus expects

  • Simplify surds using √(ab) = √a × √b and √(a/b) = √a ÷ √b
  • Add, subtract, multiply and divide expressions containing surds
  • Rationalise the denominator of fractions such as 1/√a and 1/(a + b√c)
  • Express answers in a required form such as a + b√c, where a and b are rational or integers as stated
  • Solve equations involving surds, including those that need squaring both sides, and check for invalid solutions
  • Use surds to give exact answers in problems on lengths and areas

Simplifying surds

The two rules you use most are √(ab) = √a × √b and √(a/b) = √a ÷ √b, for positive a and b. To simplify a surd, look for the largest perfect square factor: √72 = √(36 × 2) = 6√2, and √50 = √(25 × 2) = 5√2. Like surds can be added just like algebraic terms, so 6√2 + 5√2 = 11√2, but √2 + √3 cannot be combined. Note that √a + √b is not equal to √(a + b). For example, √9 + √16 = 3 + 4 = 7, while √25 = 5. When multiplying, treat surds like letters and expand brackets carefully: (3 + √2)(1 - 2√2) = 3 - 6√2 + √2 - 2(2) = -1 - 5√2, since √2 × √2 = 2.

Rationalising the denominator

An answer is not in its simplest form if it has a surd in the denominator. For a single surd, multiply the top and bottom by that surd: 6/√3 = 6√3/3 = 2√3. For a denominator such as a + √b, multiply the top and bottom by its conjugate a - √b. This works because (a + √b)(a - √b) = a² - b, which has no surd. For example, 4/(3 + √5) = 4(3 - √5)/((3 + √5)(3 - √5)) = 4(3 - √5)/(9 - 5) = 3 - √5. The same idea works for √a + √b, whose conjugate is √a - √b, and for denominators like 2√3 - 1, whose conjugate is 2√3 + 1. Always simplify after rationalising by cancelling any common factor.

Equations with surds

Two kinds of equation appear. In the first, the unknowns are rational numbers attached to surds. To find integers a and b with (a + b√3)² = 13 + 4√3, expand to get a² + 3b² + 2ab√3. Comparing the rational parts and the surd parts gives a² + 3b² = 13 and 2ab = 4. Trying integer pairs with ab = 2 gives a = 1 and b = 2, or a = -1 and b = -2. In the second kind, the unknown sits inside a square root, such as √(2x + 3) = x. Square both sides to get 2x + 3 = x², so x² - 2x - 3 = 0 and x = 3 or x = -1. Squaring can create false solutions, so substitute back: x = 3 gives √9 = 3, which works, but x = -1 gives √1 = 1, which is not -1. So x = 3 only.

Surds in lengths and areas

Many A-Math questions give lengths in surd form. A rectangle with sides (2 + √3) cm and (2 - √3) cm has area 4 - 3 = 1 cm². A right-angled triangle with shorter sides (√2 + 1) cm and (√2 - 1) cm has hypotenuse squared equal to (3 + 2√2) + (3 - 2√2) = 6, so the hypotenuse is √6 cm. When you need an unknown side, divide and then rationalise. A rectangle of area (7 + 3√5) cm² and width (3 + √5) cm has length (7 + 3√5)/(3 + √5). Multiplying the top and bottom by 3 - √5 gives (21 - 7√5 + 9√5 - 15)/(9 - 5) = (6 + 2√5)/4 = (3 + √5)/2 cm. You can check by multiplying the length by the width to get back the area.

Worked example

Without using a calculator, express (5 + 2√3)/(2 - √3) in the form a + b√3, where a and b are integers.

  1. The conjugate of the denominator 2 - √3 is 2 + √3, so multiply the numerator and the denominator by 2 + √3.
  2. Denominator: (2 - √3)(2 + √3) = 4 - 3 = 1.
  3. Numerator: (5 + 2√3)(2 + √3) = 10 + 5√3 + 4√3 + 2√3 × √3.
  4. Since √3 × √3 = 3, the last term is 2 × 3 = 6, so the numerator is 16 + 9√3.
  5. Divide by the denominator, which is 1: the result is 16 + 9√3.
  6. Check: (16 + 9√3)(2 - √3) = 32 - 16√3 + 18√3 - 27 = 5 + 2√3, which matches the original numerator.

Answer: 16 + 9√3, so a = 16 and b = 9

Common mistakes

  • Writing √(a + b) = √a + √b. Test it with numbers: √(9 + 16) = 5, but √9 + √16 = 7.
  • Forgetting that √3 × √3 = 3 when expanding, and leaving terms such as 2√3 × √3 unsimplified instead of writing 6.
  • Multiplying only the denominator by the conjugate. Whatever you multiply the bottom by, you must also multiply the top by, or the value changes.
  • Not checking solutions of surd equations after squaring. Squaring both sides can introduce a value that makes the original equation false, so substitute each answer back.
  • Simplifying √48 as 2√12 and stopping. Always take out the largest square factor, so √48 = √(16 × 3) = 4√3.
  • Giving a calculator decimal when the question says 'without using a calculator' or asks for an exact value. Leave the answer in surd form.

Exam tips

  • During practice, do a quick decimal check on a calculator to confirm that your surd answer has the same value as the original expression. In the exam, only do this when calculators are allowed.
  • When asked for a + b√c, compare rational parts and surd parts separately. One equation with surds gives you two equations without them.
  • Write √3 × √3 = 3 as a separate line when you expand brackets. It is an easy place to drop a mark.

Practice questions

1. Simplify √75 + √27 - √12.

Show answer

6√3. Each term simplifies: √75 = 5√3, √27 = 3√3 and √12 = 2√3, so the total is 5√3 + 3√3 - 2√3 = 6√3.

2. Rationalise the denominator of (√5 + 1)/(√5 - 1), giving your answer in the form (a + √5)/b, where a and b are integers.

Show answer

(3 + √5)/2. Multiply the top and bottom by √5 + 1: the numerator is (√5 + 1)² = 5 + 2√5 + 1 = 6 + 2√5 and the denominator is 5 - 1 = 4. So the fraction is (6 + 2√5)/4 = (3 + √5)/2.

3. Solve the equation √(3x + 1) = x - 3.

Show answer

x = 8. Squaring gives 3x + 1 = x² - 6x + 9, so x² - 9x + 8 = 0 and (x - 1)(x - 8) = 0. Check x = 8: √25 = 5 and 8 - 3 = 5, which works. Check x = 1: √4 = 2 but 1 - 3 = -2, so x = 1 is rejected.

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