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Surds

What you'll learn

  • Recognise a surd and simplify square roots into exact form.
  • Expand brackets containing surds.
  • Rationalise denominators so there are no roots on the bottom.
  • Use the same ideas with simple algebraic surds.

Square roots and surds

A square number is made by multiplying an integer by itself. For example, 36 is a square number because 6 times 6 is 36.

An integer is a whole number: it can be positive, negative or zero.

A rational number can be written as a fraction of integers. An irrational number cannot be written exactly as a fraction.

Definition

Surd

A surd is an irrational root left in exact form, such as 2\sqrt{2}2​ or 5\sqrt{5}5​. A root like 36\sqrt{36}36​ is not a surd because it simplifies to 6.

Simplifying surds

To simplify a surd, look for the largest square number that is a factor.

Key Idea

Simplest surd form

Factor out the largest square number first. This gets you to the simplest form quickly and avoids doing extra work.

Example

Writing 72​ in the form k2​

Factorising 72 into its largest square factor shows why the root simplifies cleanly.

  1. Find the largest square factor of 72. Since 72 is 36 times 2, write:

    72=36×2\sqrt{72}=\sqrt{36 \times 2}72​=36×2​

  2. Split the square root into two parts:

    36×2=362\sqrt{36 \times 2}=\sqrt{36}\sqrt{2}36×2​=36​2​

  3. Simplify 36\sqrt{36}36​:

    72=62\sqrt{72}=6\sqrt{2}72​=62​
Common Mistake

Splitting addition

You may split multiplication inside a square root, but not addition. For example, 9+16≠9+16\sqrt{9+16}\neq \sqrt{9}+\sqrt{16}9+16​=9​+16​.

Surds with coefficients

A coefficient is the number multiplying an expression. In 535\sqrt{3}53​, the coefficient is 5.

When there is already a number in front of the surd, simplify the root first, then multiply the coefficients.

Example

Writing 545​ in the form k5​

The coefficient stays outside while the root part is simplified first.

  1. Simplify the square root part:

    45=9×5=35\sqrt{45}=\sqrt{9 \times 5}=3\sqrt{5}45​=9×5​=35​

  2. Put this back into the expression:

    545=5×355\sqrt{45}=5 \times 3\sqrt{5}545​=5×35​

  3. Multiply the coefficients:

    545=1555\sqrt{45}=15\sqrt{5}545​=155​

Expanding brackets with surds

To expand means to multiply out the brackets. Treat surds like algebra terms, but remember that a surd times itself becomes a whole number, for example 7×7=7\sqrt{7}\times\sqrt{7}=77​×7​=7.

Like surds have the same root part, such as 323\sqrt{2}32​ and −52-5\sqrt{2}−52​. You can add or subtract like surds.

Example

Expanding (3+2​)(2−2​)

A grid helps make sure all four products are included when expanding brackets with surds.

  1. Multiply each term in the first bracket by each term in the second bracket:

    (3+2)(2−2)(3+\sqrt{2})(2-\sqrt{2})(3+2​)(2−2​)

  2. Write out the four products:

    6−32+22−26-3\sqrt{2}+2\sqrt{2}-26−32​+22​−2

  3. Combine the number parts and the surd parts:

    4−24-\sqrt{2}4−2​

Squaring a bracket

Squaring a bracket means multiplying the bracket by itself. Do not just square the two terms separately — there is usually a middle term.

Example

Writing (5−3​)2 in the form a+b3​

Writing the square as two brackets makes the two middle terms visible.

  1. Rewrite the square as two brackets:

    (5−3)2=(5−3)(5−3)(5-\sqrt{3})^2=(5-\sqrt{3})(5-\sqrt{3})(5−3​)2=(5−3​)(5−3​)

  2. Expand carefully:

    25−53−53+325-5\sqrt{3}-5\sqrt{3}+325−53​−53​+3

  3. Collect like terms:

    28−10328-10\sqrt{3}28−103​
Common Mistake

Forgetting the middle terms

The expression (5−3)2(5-\sqrt{3})^2(5−3​)2 is not 25+325+325+3. The two middle terms, −53-5\sqrt{3}−53​ and −53-5\sqrt{3}−53​, must be included.

Conjugates

Definition

Conjugates

Conjugates are two expressions that differ only by the sign between the terms, such as 4+74+\sqrt{7}4+7​ and 4−74-\sqrt{7}4−7​.

Conjugates are useful because the surd parts cancel when multiplied:

Multiplying conjugates cancels the surd terms and leaves a difference of two squares.

(a+b)(a−b)=a2−b(a+\sqrt{b})(a-\sqrt{b})=a^2-b(a+b​)(a−b​)=a2−b
Example

Expanding (7+23​)(7−23​)

  1. Notice that the brackets are conjugates, so use difference of two squares:

    (7+23)(7−23)=72−(23)2(7+2\sqrt{3})(7-2\sqrt{3})=7^2-(2\sqrt{3})^2(7+23​)(7−23​)=72−(23​)2

  2. Square each part:

    49−(4×3)49-(4 \times 3)49−(4×3)

  3. Simplify:

    373737

Rationalising denominators

Definition

Rationalising the denominator

To rationalise the denominator means to rewrite a fraction so that there is no surd on the bottom.

If the denominator is a single surd, multiply the top and bottom by that surd.

Example

Simplifying 2​4+8​​

Multiplying by the same surd over itself removes the surd from a single-term denominator.

  1. Multiply the numerator and denominator by 2\sqrt{2}2​:

    4+82×22\frac{4+\sqrt{8}}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}2​4+8​​×2​2​​

  2. Expand the numerator and simplify the denominator:

    42+162\frac{4\sqrt{2}+\sqrt{16}}{2}242​+16​​

  3. Simplify fully:

    42+42=22+2\frac{4\sqrt{2}+4}{2}=2\sqrt{2}+2242​+4​=22​+2

Rationalising with a conjugate

If the denominator has two terms, such as 2+32+\sqrt{3}2+3​, multiply by its conjugate, 2−32-\sqrt{3}2−3​.

Example

Showing 2+3​4+3​​=5−23​

For a two-term denominator, multiplying by the conjugate makes the surd parts cancel.

  1. Multiply the top and bottom by the conjugate of the denominator:

    4+32+3×2−32−3\frac{4+\sqrt{3}}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}2+3​4+3​​×2−3​2−3​​

  2. Expand the denominator:

    (2+3)(2−3)=4−3=1(2+\sqrt{3})(2-\sqrt{3})=4-3=1(2+3​)(2−3​)=4−3=1

  3. Expand the numerator:

    (4+3)(2−3)=8−43+23−3(4+\sqrt{3})(2-\sqrt{3})=8-4\sqrt{3}+2\sqrt{3}-3(4+3​)(2−3​)=8−43​+23​−3

  4. Simplify:

    5−235-2\sqrt{3}5−23​
Tip

Choosing what to multiply by

For a two-term denominator, change only the sign in the middle. The conjugate of 3−53-\sqrt{5}3−5​ is 3+53+\sqrt{5}3+5​.

Fractions inside fractions

Sometimes the denominator contains a small fraction. First combine the denominator into one fraction, then simplify.

Example

Simplifying 5​1​+5​1​

Combining the denominator into one fraction first makes the complex fraction easier to simplify.

  1. Write both parts of the denominator over 5\sqrt{5}5​:

    15+5=15+55\frac{1}{\sqrt{5}}+\sqrt{5}=\frac{1}{\sqrt{5}}+\frac{5}{\sqrt{5}}5​1​+5​=5​1​+5​5​

  2. Add the fractions in the denominator:

    15+5=65\frac{1}{\sqrt{5}}+\sqrt{5}=\frac{6}{\sqrt{5}}5​1​+5​=5​6​

  3. Divide by a fraction by multiplying by its reciprocal:

    165=56\frac{1}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{5}}{6}5​6​1​=65​​

Algebraic surds

The same rules work with letters. For GCSE questions, assume the expressions under square roots are non-negative unless told otherwise.

Common Mistake

Variables under roots

A square root such as x\sqrt{x}x​ only makes sense in GCSE real-number work when x≥0x\ge 0x≥0. Also, denominators must not be zero.

Example

Simplifying algebraic surds

  1. Use conjugates to simplify (m+n)(m−n)(\sqrt{m}+\sqrt{n})(\sqrt{m}-\sqrt{n})(m​+n​)(m​−n​):

    (m+n)(m−n)=m−n(\sqrt{m}+\sqrt{n})(\sqrt{m}-\sqrt{n})=m-n(m​+n​)(m​−n​)=m−n

  2. Expand (3p+q)2(3p+\sqrt{q})^2(3p+q​)2 by writing it as two brackets:

    (3p+q)2=(3p+q)(3p+q)(3p+\sqrt{q})^2=(3p+\sqrt{q})(3p+\sqrt{q})(3p+q​)2=(3p+q​)(3p+q​)

  3. Multiply out and collect terms:

    9p2+6pq+q9p^2+6p\sqrt{q}+q9p2+6pq​+q
Exam technique

In the exam

  1. Look for square factors first when simplifying a single surd.
  2. When expanding brackets, write all four products before collecting terms.
  3. To rationalise a two-term denominator, multiply by the conjugate and simplify carefully.
Self review

Check yourself

  • Can you simplify a surd by finding the largest square factor?
  • Can you expand a squared bracket without losing the middle terms?
  • Can you choose the correct conjugate to rationalise a denominator?

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

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Surds Revision Guide

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