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Maths · Further Algebra

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Surds

Simplifying and calculating with surds, expanding brackets and rationalising denominators.

  • Higher
  • 9 key terms
  • All boards

Learning Objectives

  1. 1Recognise a surd and explain why it is left as a root rather than a decimal.
  2. 2Simplify surds by taking out square factors, such as \(\sqrt{12} = 2\sqrt{3}\).
  3. 3Add, subtract, multiply and divide surds, and expand brackets containing surds.
  4. 4Rationalise a denominator, including one of the form \(a + \sqrt{b}\).

Exact answers

A surd is a root that cannot be written as a whole number or a fraction, such as \(\sqrt{2}\) or \(\sqrt{5}\). Its decimal never ends, so on a non-calculator paper you leave it as a root, which is also exact. Pythagoras and trigonometry produce surds all the time, so this lesson is the algebra that tidies those answers up. Surds are Higher tier content, and every method here rests on one rule: \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\).

Simplifying surds

Look for the biggest square number that divides into the number under the root.

  • Method

    Split the number into a square number times another number, then take the root of the square. \(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\).

  • Squares to know

    4, 9, 16, 25, 36, 49.

  • Examples

    \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\), and \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\).

  • Simplest form

    Use the largest square factor, so \(\sqrt{48} = 4\sqrt{3}\), not \(2\sqrt{12}\).

Simplifying a surd

Write \(\sqrt{72}\) in the form \(a\sqrt{2}\).

Show the solutionHide the solution
  1. 1 Find a square factor \(72 = 36 \times 2\).
  2. 2 Split the root \(\sqrt{72} = \sqrt{36} \times \sqrt{2}\).
  3. 3 Take the square root of the square \(\sqrt{36} = 6\).
  4. 4 Write the answer \(6\sqrt{2}\).

Answer\(6\sqrt{2}\)

Adding, subtracting and multiplying

Surds behave like letters when they add and like numbers when they multiply.

  • Like surds add

    \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\), just as \(3x + 5x = 8x\).

  • Simplify first

    \(\sqrt{8} + \sqrt{2} = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2}\).

  • Unlike surds cannot be added

    \(\sqrt{2} + \sqrt{3}\) stays as it is.

  • Multiplying

    \(\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4\), and \(3\sqrt{2} \times 2\sqrt{5} = 6\sqrt{10}\).

Expanding brackets with surds

Use the same methods as with algebra, then simplify.

  • Single bracket

    \(\sqrt{2}(3 + \sqrt{2}) = 3\sqrt{2} + 2\).

  • Double brackets

    \((1 + \sqrt{3})(2 - \sqrt{3}) = 2 - \sqrt{3} + 2\sqrt{3} - 3 = -1 + \sqrt{3}\).

  • Squares

    \((2 + \sqrt{5})^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5}\).

  • Difference of two squares

    \((3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7\), and the surd disappears.

Rationalising the denominator

Leaving a surd on the bottom of a fraction is untidy, so multiply to remove it.

  • A single surd

    Multiply top and bottom by that surd. \(\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}\).

  • A two-part denominator

    Multiply by the conjugate, the same two terms with the sign changed. For \(2 + \sqrt{3}\) use \(2 - \sqrt{3}\).

  • Why it works

    \((2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1\), so the bottom becomes a whole number.

  • Example

    \(\dfrac{1}{2 + \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = 2 - \sqrt{3}\).

Rationalising

Rationalise the denominator of \(\dfrac{10}{\sqrt{5}}\), and simplify your answer.

Show the solutionHide the solution
  1. 1 Multiply top and bottom by the surd \(\dfrac{10}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{10\sqrt{5}}{5}\).
  2. 2 Cancel \(\dfrac{10\sqrt{5}}{5} = 2\sqrt{5}\).
  3. 3 Check \(\dfrac{\sqrt{5}}{\sqrt{5}} = 1\), so multiplying by it changes the look of the fraction but not its value.

Answer\(2\sqrt{5}\)

Test yourself

  1. 1

    What is \(\sqrt{5} \times \sqrt{5}\)?

    Show answerHide answer

    5.

  2. 2

    Simplify \(\sqrt{12}\).

    Show answerHide answer

    \(2\sqrt{3}\).

  3. 3

    What is \(\sqrt{2} \times \sqrt{8}\)?

    Show answerHide answer

    \(\sqrt{16} = 4\).

  4. 4

    What is the conjugate of \(3 + \sqrt{2}\)?

    Show answerHide answer

    \(3 - \sqrt{2}\).

  5. 5

    What is \((3 + \sqrt{2})(3 - \sqrt{2})\)?

    Show answerHide answer

    7.

Exam technique: surds

These questions are about method and tidiness.

  • Simplify as you go

    Taking out square factors early makes later steps easier.

  • Leave answers exact

    Do not give a decimal unless the question asks for it.

  • Show the rationalising line

    Writing the multiplication earns the method mark.

  • Check the form

    "In the form \(a\sqrt{b}\)" means one surd with a whole number in front.

Summary and exam focus

  • A surd is a root that is not a whole number, left exact as \(\sqrt{n}\).
  • Simplify with the biggest square factor: \(\sqrt{50} = 5\sqrt{2}\).
  • Multiply surds with \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\), and add only like surds.
  • Rationalise a single surd by multiplying by it, and \(a + \sqrt{b}\) by its conjugate.

Exam focus

Write \(\sqrt{48} + \sqrt{3}\) in the form \(a\sqrt{3}\), where \(a\) is an integer. (2 marks) (2 marks)

Simplify \(\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\) first, and then add \(\sqrt{3}\) to get \(5\sqrt{3}\). Only like surds can be added, and you make them alike by simplifying.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Surd
A root of a number that cannot be written as a whole number or fraction.
Irrational
A number that cannot be written as a fraction, with a decimal that never ends or repeats.
Rationalise
Remove a surd from the denominator of a fraction.
Conjugate
An expression with the sign between two terms changed, such as \(2 - \sqrt{3}\) for \(2 + \sqrt{3}\).
Square factor
A factor that is a square number, such as 4 or 9.
Like surds
Surds with the same number under the root, which can be added or subtracted.
Exact value
An answer left in terms of surds or \(\pi\) rather than rounded.
Square root
The number that gives a stated number when multiplied by itself.
Denominator
The bottom of a fraction.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Simplify 2 marks Core

Write \(\sqrt{45}\) in the form \(a\sqrt{5}\), where \(a\) is an integer.

Mark scheme — 2 marks available

  • \(\sqrt{9 \times 5}\) — M1
  • \(3\sqrt{5}\) — A1

Model answer

\(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\).

2. Exam question Expand 3 marks Core

(a) Expand and simplify \((\sqrt{5} + 2)^2\). [2 marks] (b) Show that \((\sqrt{5} + 2)(\sqrt{5} - 2) = 1\). [1 mark]

Mark scheme — 3 marks available

  • (a) Three of the four terms correct — M1
  • (a) \(9 + 4\sqrt{5}\) — A1
  • (b) \(5 - 4 = 1\) shown — Q1

Model answer

(a) \(5 + 2\sqrt{5} + 2\sqrt{5} + 4 = 9 + 4\sqrt{5}\). (b) \(5 - 2\sqrt{5} + 2\sqrt{5} - 4 = 1\).

3. Exam question Simplify 3 marks Core

Write \(\sqrt{200} - \sqrt{50}\) in the form \(a\sqrt{2}\).

Mark scheme — 3 marks available

  • \(10\sqrt{2}\) or \(5\sqrt{2}\) seen — M1
  • Both correct — M1
  • \(5\sqrt{2}\) — A1

Model answer

\(\sqrt{200} = 10\sqrt{2}\) and \(\sqrt{50} = 5\sqrt{2}\), so the difference is \(5\sqrt{2}\).

4. Exam question Rationalise 2 marks Core

Rationalise the denominator of \(\dfrac{10}{\sqrt{5}}\).

Mark scheme — 2 marks available

  • Multiplies the top and bottom by \(\sqrt{5}\) — M1
  • \(2\sqrt{5}\) — A1

Model answer

\(\dfrac{10\sqrt{5}}{5} = 2\sqrt{5}\).

5. Exam question Show that 3 marks Stretch

Show that \(\dfrac{4}{\sqrt{7} - \sqrt{3}} = \sqrt{7} + \sqrt{3}\).

Mark scheme — 3 marks available

  • Multiplies by \(\sqrt{7} + \sqrt{3}\) — M1
  • Denominator \(7 - 3 = 4\) — M1
  • Completes with a conclusion — Q1

Model answer

Multiply the top and bottom by \(\sqrt{7} + \sqrt{3}\): the bottom becomes \(7 - 3 = 4\), and the fraction is \(\dfrac{4(\sqrt{7} + \sqrt{3})}{4} = \sqrt{7} + \sqrt{3}\).

6. Exam question Work out 4 marks Stretch

A rectangle has length \((3 + \sqrt{2})\) cm and width \((3 - \sqrt{2})\) cm. (a) Show that the area of the rectangle is 7 cm\(^2\). [2 marks] (b) Work out the perimeter of the rectangle. [2 marks]

Mark scheme — 4 marks available

  • (a) \((3 + \sqrt{2})(3 - \sqrt{2})\) expanded — M1
  • (a) \(9 - 2 = 7\) — Q1
  • (b) \(2(3 + \sqrt{2} + 3 - \sqrt{2})\) — M1
  • (b) 12 cm — A1

Model answer

(a) \((3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7\). (b) The perimeter is \(2 \times ((3 + \sqrt{2}) + (3 - \sqrt{2})) = 2 \times 6 = 12\) cm.

7. Multiple choice 1 mark Easier

What is \(\sqrt{5} \times \sqrt{5}\)?

  1. A \(5\) Correct
  2. B \(\sqrt{10}\)
  3. C \(25\)
  4. D \(2\sqrt{5}\)

Why: A root times itself gives the number: \(\sqrt{a} \times \sqrt{a} = a\).

8. Multiple choice 1 mark Core

Simplify \(\sqrt{12}\).

  1. A \(3\sqrt{2}\)
  2. B \(6\)
  3. C \(4\sqrt{3}\)
  4. D \(2\sqrt{3}\) Correct

Why: \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\).

9. Multiple choice 1 mark Core

Work out \(\sqrt{2} \times \sqrt{8}\).

  1. A \(\sqrt{10}\)
  2. B \(8\)
  3. C \(4\) Correct
  4. D \(2\sqrt{2}\)

Why: \(\sqrt{2 \times 8} = \sqrt{16} = 4\).

10. Multiple choice 1 mark Easier

Work out \(3\sqrt{2} + 5\sqrt{2}\).

  1. A \(15\sqrt{2}\)
  2. B \(8\sqrt{2}\) Correct
  3. C \(8\sqrt{4}\)
  4. D \(8\)

Why: Like surds add, as in \(3x + 5x = 8x\).

11. Multiple choice 1 mark Core

Simplify \(\sqrt{50}\).

  1. A \(5\sqrt{2}\) Correct
  2. B \(25\sqrt{2}\)
  3. C \(10\sqrt{5}\)
  4. D \(2\sqrt{5}\)

Why: \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\).

12. Multiple choice 1 mark Core

Rationalise the denominator of \(\dfrac{6}{\sqrt{3}}\).

  1. A \(6\sqrt{3}\)
  2. B \(2\)
  3. C \(\sqrt{3}\)
  4. D \(2\sqrt{3}\) Correct

Why: \(\dfrac{6\sqrt{3}}{3} = 2\sqrt{3}\).

13. Multiple choice 1 mark Core

Expand and simplify \((3 + \sqrt{2})(3 - \sqrt{2})\).

  1. A \(11\)
  2. B \(9\)
  3. C \(7\) Correct
  4. D \(9 - 2\sqrt{2}\)

Why: This is a difference of two squares: \(9 - 2 = 7\).

14. Multiple choice 1 mark Stretch

Write \(\sqrt{48} + \sqrt{3}\) in the form \(a\sqrt{3}\).

  1. A \(7\sqrt{3}\)
  2. B \(5\sqrt{3}\) Correct
  3. C \(\sqrt{51}\)
  4. D \(4\sqrt{6}\)

Why: \(\sqrt{48} = 4\sqrt{3}\), and \(4\sqrt{3} + \sqrt{3} = 5\sqrt{3}\).

15. Multiple choice 1 mark Stretch

Rationalise the denominator of \(\dfrac{1}{2 + \sqrt{3}}\).

  1. A \(2 - \sqrt{3}\) Correct
  2. B \(2 + \sqrt{3}\)
  3. C \(\dfrac{1}{2}\)
  4. D \(\dfrac{2 - \sqrt{3}}{7}\)

Why: Multiply top and bottom by \(2 - \sqrt{3}\); the bottom becomes \(4 - 3 = 1\).