This Higher tier note covers surds: numbers such as \(\sqrt{2}\) left in root form so that answers stay exact. You will learn to simplify surds, add, subtract and multiply them, and rationalise denominators. Surds appear in trigonometry, Pythagoras' theorem and the quadratic formula.
Key ideas
A surd is a root that cannot be written as a fraction of whole numbers, such as \(\sqrt{2}\), \(\sqrt{7}\) or \(\sqrt{10}\). Its decimal never ends or repeats. In contrast, \(\sqrt{16}=4\) is not a surd. Leaving an answer as a surd gives an exact value, while a decimal is only an approximation. When a question asks for an exact answer, do not use a calculator decimal.
Two key rules for square roots, for \(a\ge0\) and \(b\gt0\):
- \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\)
- \(\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}\)
Also \(\sqrt{a}\times\sqrt{a}=a\). There is no rule for \(\sqrt{a+b}\); it is not equal to \(\sqrt{a}+\sqrt{b}\).
Simplifying. Find the largest square number that divides the number under the root, then split the root. For example, \(\sqrt{72}=\sqrt{36\times2}=\sqrt{36}\times\sqrt{2}=6\sqrt{2}\). A surd is in simplest form when the number under the root has no square factor except 1.
Adding and subtracting. Only like surds can be combined, which means surds with the same number under the root. First simplify each surd, then collect terms as you would with algebra, treating \(\sqrt{3}\) like a letter. For example, \(2\sqrt{3}+5\sqrt{3}=7\sqrt{3}\), but \(\sqrt{2}+\sqrt{3}\) cannot be simplified.
Expanding brackets. Multiply as with algebra. Use \(\sqrt{a}\times\sqrt{a}=a\) to turn squares of surds into whole numbers. Notice that \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\) is always a rational number, because the surd terms cancel.
Rationalising the denominator. This means rewriting a fraction so that the denominator is a whole number.
- If the denominator is a single surd such as \(\sqrt{k}\), multiply top and bottom by \(\sqrt{k}\).
- If the denominator is of the form \(a+\sqrt{b}\), multiply top and bottom by the conjugate \(a-\sqrt{b}\) (and the other way round). The denominator becomes \(a^2-b\).
Rational and irrational numbers. A rational number can be written as a fraction of two integers, so \(0.75\), \(\dfrac{2}{3}\) and \(\sqrt{25}\) are all rational. An irrational number cannot, and a surd is an irrational number written as a root. When you add a rational number to a surd, the result is irrational. For example, \(3+\sqrt{2}\) is irrational. A product of two surds, however, can be rational, as in \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\).
Surds in geometry. Pythagoras' theorem often produces surds. If the sides are whole numbers, the hypotenuse is \(\sqrt{a^2+b^2}\), which is usually a surd. Leaving an area or length as a surd avoids rounding errors that could build up in later steps. The area of a right-angled triangle with perpendicular sides \(\sqrt{6}\) cm and \(\sqrt{3}\) cm is \(\dfrac12\times\sqrt{6}\times\sqrt{3}=\dfrac12\sqrt{18}=\dfrac{3\sqrt{2}}{2}\) cm\(^2\).
Answer forms. Questions often ask for the answer in a given form, such as \(a\sqrt{b}\) or \(a+b\sqrt{c}\), where \(a\), \(b\) and \(c\) are integers or fractions. After expanding or rationalising, collect the whole-number terms and the surd terms separately. A final check on the calculator, comparing the decimal value of the start and end of your working, catches most slips.
Checking by estimation. You can check a surd answer quickly by remembering that \(\sqrt{2}\approx1.41\), \(\sqrt{3}\approx1.73\) and \(\sqrt{5}\approx2.24\). For example, \(7\sqrt{3}\) should be a little over 12, which agrees with a direct calculation of \(\sqrt{48}+\sqrt{75}-\sqrt{12}\approx6.93+8.66-3.46=12.1\).
Why rationalise? A fraction with a whole-number denominator is easier to compare, add and estimate, and it gives a standard form that is easy to recognise as correct. For example, \(\dfrac{1}{\sqrt{2}}\) and \(\dfrac{\sqrt{2}}{2}\) are equal, but the second is the preferred form. Always check whether the result can be simplified further after you rationalise.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Surd | An irrational root that is left unevaluated, such as \(\sqrt{5}\). |
| \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) | Multiplying or splitting a root to simplify. |
| \(\sqrt{a}\div\sqrt{b}=\sqrt{a/b}\) | Dividing roots. |
| \(\sqrt{a}\times\sqrt{a}=a\) | Squaring a surd gives a whole number. |
| \(m\sqrt{a}\pm n\sqrt{a}=(m\pm n)\sqrt{a}\) | Combining like surds. |
| \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\) | Difference of two squares; used to rationalise. |
| Conjugate | The same expression with the sign of the surd term changed. |
| Rationalising | Removing surds from the denominator. |
Worked examples
Example 1 Simplify \(\sqrt{48}+\sqrt{75}-\sqrt{12}\), giving your answer in the form \(a\sqrt{b}\).
Simplify each term using the largest square factor.
\(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\).
\(\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}\).
\(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\).
Now all three are like surds: \(4\sqrt{3}+5\sqrt{3}-2\sqrt{3}=7\sqrt{3}\).
Answer: \(7\sqrt{3}\).
Example 2 Expand and simplify \((3+\sqrt{5})(2-\sqrt{5})\). Then find the exact length of the diagonal of a rectangle that is 5 cm by 3 cm.
Multiply each term in the first bracket by each term in the second: \(3\times2=6\), \(3\times(-\sqrt{5})=-3\sqrt{5}\), \(\sqrt{5}\times2=2\sqrt{5}\), \(\sqrt{5}\times(-\sqrt{5})=-5\).
Add: \(6-3\sqrt{5}+2\sqrt{5}-5=1-\sqrt{5}\).
For the diagonal, use Pythagoras' theorem: \(d^2=5^2+3^2=34\), so \(d=\sqrt{34}\). Since 34 \(=2\times17\) has no square factor, this is already simplified.
Answer: \(1-\sqrt{5}\); the diagonal is \(\sqrt{34}\) cm (about 5.83 cm).
Example 3 Rationalise the denominator of (a) \(\dfrac{6}{\sqrt{3}}\) and (b) \(\dfrac{10}{4-\sqrt{6}}\).
(a) Multiply top and bottom by \(\sqrt{3}\): \(\dfrac{6\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\dfrac{6\sqrt{3}}{3}=2\sqrt{3}\).
(b) The conjugate of \(4-\sqrt{6}\) is \(4+\sqrt{6}\). Multiply top and bottom by it.
Denominator: \((4-\sqrt{6})(4+\sqrt{6})=16-6=10\).
Numerator: \(10(4+\sqrt{6})\).
So the fraction is \(\dfrac{10(4+\sqrt{6})}{10}=4+\sqrt{6}\).
Answer: (a) \(2\sqrt{3}\); (b) \(4+\sqrt{6}\).
Common errors
- Writing \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\). For example, \(\sqrt{9+16}=5\), not \(3+4=7\).
- Adding unlike surds, such as \(\sqrt{2}+\sqrt{8}\), without first simplifying. In fact \(\sqrt{8}=2\sqrt{2}\), so the sum is \(3\sqrt{2}\).
- Stopping at a surd that can still be simplified, such as leaving \(\sqrt{50}\) instead of \(5\sqrt{2}\).
- When squaring a bracket such as \((2+\sqrt{3})^2\), writing only \(4+3\). The middle term \(4\sqrt{3}\) is needed.
- Rationalising with the wrong sign, which does not remove the surd from the denominator.
Exam tip: when the question says "exact" or "in the form \(a+b\sqrt{c}\)", every line of working should keep surds. Avoid decimals until you check your answer with a calculator at the end.
Check yourself
- Write \(\sqrt{50}\) in the form \(a\sqrt{2}\).
Answer
\(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). - Expand and simplify \((\sqrt{5}+2)^2\).
Answer
\((\sqrt{5}+2)(\sqrt{5}+2)=5+2\sqrt{5}+2\sqrt{5}+4=9+4\sqrt{5}\). - Rationalise \(\dfrac{14}{3-\sqrt{2}}\).
Answer
Multiply by \(\dfrac{3+\sqrt{2}}{3+\sqrt{2}}\). Denominator: \(9-2=7\). Numerator: \(14(3+\sqrt{2})\). So the result is \(2(3+\sqrt{2})=6+2\sqrt{2}\). - Simplify \(\sqrt{18}\times\sqrt{8}\).
Answer
\(\sqrt{18\times8}=\sqrt{144}=12\). Alternatively, \(3\sqrt{2}\times2\sqrt{2}=6\times2=12\).