Multiplying and Dividing Surds
Every question with a fully worked solution.
- Multiplying and Dividing Surds - Video - Multiplying Surds Watch
Theory
To multiply surds, multiply the numbers outside together and the numbers inside together, then simplify: \(3\sqrt{2}\times 4\sqrt{5} = 12\sqrt{10}\). Dividing works the same way. This Year 10A maths page covers signs, powers of surds, pronumerals under the root and exact areas and volumes.
For \(a, b \ge 0\), \(\sqrt{a}\times\sqrt{b} = \sqrt{ab}\) and, for \(b \gt 0\), \(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}\). Unlike adding, you do not need like surds to multiply or divide.
With numbers in front, treat them like coefficients: \(m\sqrt{a}\times n\sqrt{b} = mn\sqrt{ab}\). Always simplify the surd at the end, because the product often contains a square factor.
The key fact is \(\sqrt{a}\times\sqrt{a} = (\sqrt{a})^2 = a\). So a surd squared is rational, and higher powers split into squares: \((\sqrt{5})^3 = (\sqrt{5})^2\times\sqrt{5} = 5\sqrt{5}\).
If a division leaves a surd in the denominator, multiply the top and bottom by that surd. This rationalises the denominator without changing the value.
| Power | Rule | Example |
|---|---|---|
| Square | \((\sqrt{a})^2 = a\) | \((\sqrt{11})^2 = 11\) |
| Square with a coefficient | \((k\sqrt{a})^2 = k^2a\) | \((4\sqrt{5})^2 = 80\) |
| Cube | \((\sqrt{a})^3 = a\sqrt{a}\) | \((\sqrt{7})^3 = 7\sqrt{7}\) |
| Fourth power | \((\sqrt{a})^4 = a^2\) | \((\sqrt{5})^4 = 25\) |
How to multiply or divide surds
- Multiply (or divide) the numbers outside the roots.
- Multiply (or divide) the numbers inside the roots, under one root sign.
- Simplify the surd using its largest square factor, and combine with the outside number.
- Signs follow the usual rules: negative \(\times\) negative is positive.
- Rationalise if a surd is left in a denominator: multiply top and bottom by that surd.
Other question types
- Simplify first to keep numbers small: \(\sqrt{75}\times\sqrt{12}\) is \(5\sqrt{3}\times 2\sqrt{3} = 10\times 3 = 30\), with no big multiplication.
- Several factors: combine them all under one root, then look for squares.
- Find an unknown: simplify the surd side, then compare it with the other side, or solve the equation that results.
- Whole-number products: \(\sqrt{c}\times\sqrt{d}\) is a whole number exactly when \(cd\) is a perfect square. Factorise to see which factors are missing a partner.
- Measurement: area \(= l\times w\), volume \(= l\times w\times h\), and width \(=\) area \(\div\) length all use surd multiplication and division.
(i) \(5\sqrt{6}\times 2\sqrt{10}\)
(ii) \(-3\sqrt{7}\times 2\sqrt{14}\)
(i) Outside numbers together, inside numbers together, then simplify.
| \(5\sqrt{6}\times 2\sqrt{10}\) | \(=\) | \(10\sqrt{60}\) |
| \(\) | \(=\) | \(10\times 2\sqrt{15}\) |
| \(\) | \(=\) | \(\mathbf{20\sqrt{15}}\) |
(ii) The answer is negative because one factor is negative.
| \(-3\sqrt{7}\times 2\sqrt{14}\) | \(=\) | \(-6\sqrt{98}\) |
| \(\) | \(=\) | \(-6\times 7\sqrt{2}\) |
| \(\) | \(=\) | \(\mathbf{-42\sqrt{2}}\) |
Shortcut: \(\sqrt{7}\times\sqrt{14} = \sqrt{7}\times\sqrt{7}\times\sqrt{2} = 7\sqrt{2}\). Splitting into factors first keeps the numbers small.
(i) \(\dfrac{12\sqrt{30}}{4\sqrt{5}}\)
(ii) \(\dfrac{6\sqrt{5}}{4\sqrt{3}}\)
(i) Outside divided by outside, inside divided by inside.
| \(\dfrac{12\sqrt{30}}{4\sqrt{5}}\) | \(=\) | \(\dfrac{12}{4}\sqrt{\dfrac{30}{5}}\) |
| \(\) | \(=\) | \(\mathbf{3\sqrt{6}}\) |
(ii) \(5\div 3\) is not a whole number. Cancel the outside numbers, then multiply top and bottom by \(\sqrt{3}\) so the denominator is rational.
| \(\dfrac{6\sqrt{5}}{4\sqrt{3}}\) | \(=\) | \(\dfrac{3\sqrt{5}}{2\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}\) |
| \(\) | \(=\) | \(\dfrac{3\sqrt{15}}{6}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{\sqrt{15}}{2}}\) |
Multiplying by \(\dfrac{\sqrt{3}}{\sqrt{3}}\) is multiplying by \(1\), so the value does not change.
(i) \((3\sqrt{2})^3\)
(ii) \(\sqrt{12m}\times\sqrt{3m^3}\), where \(m\gt 0\)
(i) Cube both parts of the product. Use \((\sqrt{2})^2 = 2\).
| \((3\sqrt{2})^3\) | \(=\) | \(3^3\times(\sqrt{2})^3\) |
| \(\) | \(=\) | \(27\times 2\sqrt{2}\) |
| \(\) | \(=\) | \(\mathbf{54\sqrt{2}}\) |
(ii) Combine under one root, then take the square root of each factor.
| \(\sqrt{12m}\times\sqrt{3m^3}\) | \(=\) | \(\sqrt{36m^4}\) |
| \(\) | \(=\) | \(\boldsymbol{6m^2}\) |
\(\sqrt{m^4} = m^2\) because \((m^2)^2 = m^4\). The answer has no surd left.
(i) area
(ii) perimeter
(iii) diagonal length \(d\).
| \(\text{(i)}\ A\) | \(=\) | \((2\sqrt{7})^2 = 4\times 7 = \mathbf{28}\text{ cm}^2\) |
| \(\text{(ii)}\ P\) | \(=\) | \(4\times 2\sqrt{7} = \mathbf{8\sqrt{7}}\text{ cm}\) |
(iii) Pythagoras' theorem in the half-square:
| \(d^2\) | \(=\) | \(28+28 = 56\) |
| \(d\) | \(=\) | \(\sqrt{56} = \sqrt{4\times 14}\) |
| \(\) | \(=\) | \(\mathbf{2\sqrt{14}}\text{ cm}\) |
Check: a square's diagonal is \(\sqrt{2}\) times its side, and \(2\sqrt{7}\times\sqrt{2} = 2\sqrt{14}\).
Common pitfalls
Frequently asked questions
How do you multiply surds?
Multiply the numbers in front together and the numbers under the roots together, then simplify. For example \(5\sqrt{6}\times 2\sqrt{10} = 10\sqrt{60} = 20\sqrt{15}\).
What is a surd times itself?
The number under the root. \(\sqrt{7}\times\sqrt{7} = 7\), because squaring undoes the square root.
Do you need like surds to multiply?
No. Any two surds can be multiplied: \(\sqrt{2}\times\sqrt{5} = \sqrt{10}\). You only need like surds when adding or subtracting.
How do you divide surds?
Divide the numbers in front and divide the numbers under the roots. If the numbers under the roots do not divide exactly, write the answer as a fraction and rationalise the denominator.
How do you square a surd with a number in front?
Square both parts: \((4\sqrt{5})^2 = 4^2\times(\sqrt{5})^2 = 16\times 5 = 80\).