Multiplying and Dividing Surds
Surds multiply using \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\) and divide using \(\sqrt{a} \div \sqrt{b} = \sqrt{\dfrac{a}{b}}\), with whole-number parts dealt with separately. This NSW Year 9 Advanced maths subtopic builds these rules into longer calculations (NESA outcome MA5‑IND‑P‑02).
You will simplify products such as \(2\sqrt{5} \times 3\sqrt{15}\) and quotients such as \(\dfrac{6\sqrt{24}}{2\sqrt{8}}\), expand \(3\sqrt{5}(\sqrt{10} - 2\sqrt{5})\), and expand binomial products, including \((\sqrt{5} + \sqrt{2})^2\) and a pair whose product is a whole number. You will also find exact areas of shapes with surd sides and decide which of two shapes has the larger area. It is part of Year 9 Advanced Pythagoras Theorem.
Every question with a fully worked solution.
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Theory
To multiply surds, multiply the numbers in front and multiply the numbers under the roots: \(m\sqrt{a} \times n\sqrt{b} = mn\sqrt{ab}\). Dividing surds works the same way. This Year 9 Advanced maths page also covers expanding brackets with surds, perfect squares and the difference of two squares.
Surds multiply and divide under one root: \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\) and \(\sqrt{a} \div \sqrt{b} = \sqrt{a \div b}\). Unlike adding, this works for any two surds, like or unlike.
When a surd has a number in front, deal with the numbers and the roots separately: \(2\sqrt{3} \times 3\sqrt{2} = 6\sqrt{6}\). A surd times itself is the number under the root: \(\sqrt{7} \times \sqrt{7} = 7\).
Expanding brackets with surds uses the same rules as algebra. Every term in one bracket multiplies every term in the other. The special products \((a + b)^2\) and \((a + b)(a - b)\) are often quicker.
Always simplify at the end. Surds in the answer should be in simplest form, and like terms should be collected. A difference of two squares such as \((\sqrt{7} + 2)(\sqrt{7} - 2) = 3\) removes the surds completely.
For \(a, b \ge 0\):
Expanding brackets works exactly as in algebra:
How to multiply and divide surds
- Multiply or divide the numbers in front, then the numbers under the roots.
- Use \(\sqrt{a} \times \sqrt{a} = a\) whenever the same surd meets itself.
- Expand brackets term by term, or with the perfect-square and difference-of-two-squares patterns.
- Simplify every surd in the result and collect like terms: whole numbers together, each kind of surd together.
- Check a division by multiplying the answer back.
(i) \(4\sqrt{6} \times 5\sqrt{10}\)
(ii) \(18\sqrt{40} \div 6\sqrt{5}\).
(i) Multiply the numbers in front, multiply the surds, then simplify.
| \(4\sqrt{6} \times 5\sqrt{10}\) | \(=\) | \(20\sqrt{60}\) |
| \(\) | \(=\) | \(20 \times \sqrt{4} \times \sqrt{15}\) |
| \(\) | \(=\) | \(\mathbf{40\sqrt{15}}\) |
(ii) Divide the numbers in front, divide the surds, then simplify.
| \(18\sqrt{40} \div 6\sqrt{5}\) | \(=\) | \(\dfrac{18}{6}\sqrt{\dfrac{40}{5}}\) |
| \(\) | \(=\) | \(3\sqrt{8}\) |
| \(\) | \(=\) | \(3 \times 2\sqrt{2} = \mathbf{6\sqrt{2}}\) |
Multiply each term in the bracket by \(2\sqrt{3}\).
| \(\) | \(\) | \(2\sqrt{3} \times \sqrt{6} + 2\sqrt{3} \times 5\sqrt{3}\) |
| \(\) | \(=\) | \(2\sqrt{18} + 10 \times 3\) |
| \(\) | \(=\) | \(2 \times 3\sqrt{2} + 30\) |
| \(\) | \(=\) | \(\mathbf{6\sqrt{2} + 30}\) |
\(\sqrt{3} \times \sqrt{3} = 3\), so the second term is a whole number. \(6\sqrt{2}\) and \(30\) are unlike terms, so the answer has two terms.
(i) \((3 - \sqrt{5})(1 + 2\sqrt{5})\)
(ii) \((\sqrt{6} - \sqrt{2})^2\)
(iii) \((2\sqrt{5} + 3)(2\sqrt{5} - 3)\).
(i) Multiply each term by each term, then collect whole numbers and surds.
| \(\) | \(\) | \(3 + 6\sqrt{5} - \sqrt{5} - 2 \times 5\) |
| \(\) | \(=\) | \((3 - 10) + (6 - 1)\sqrt{5}\) |
| \(\) | \(=\) | \(\mathbf{-7 + 5\sqrt{5}}\) |
(ii) Use \((a - b)^2 = a^2 - 2ab + b^2\).
| \(\) | \(\) | \((\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2\) |
| \(\) | \(=\) | \(6 - 2\sqrt{12} + 2\) |
| \(\) | \(=\) | \(\mathbf{8 - 4\sqrt{3}}\) |
(iii) Use \((a + b)(a - b) = a^2 - b^2\). The surd terms cancel.
| \(\) | \(\) | \((2\sqrt{5})^2 - 3^2\) |
| \(\) | \(=\) | \(20 - 9\) |
| \(\) | \(=\) | \(\mathbf{11}\) |
(i) its exact area
(ii) the exact length of its hypotenuse.
(i) The shorter sides are the base and height. Their product is a difference of two squares.
| \(A\) | \(=\) | \(\dfrac{1}{2}(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})\) |
| \(\) | \(=\) | \(\dfrac{1}{2}(6 - 2)\) |
| \(\) | \(=\) | \(\mathbf{2}\ \text{cm}^2\) |
(ii) Square each side with the perfect-square rule. The \(\sqrt{12}\) terms cancel when added.
| \((\sqrt{6} + \sqrt{2})^2\) | \(=\) | \(8 + 2\sqrt{12}\) |
| \((\sqrt{6} - \sqrt{2})^2\) | \(=\) | \(8 - 2\sqrt{12}\) |
| \(c^2\) | \(=\) | \((8 + 2\sqrt{12}) + (8 - 2\sqrt{12})\) |
| \(\) | \(=\) | \(16\) |
| \(c\) | \(=\) | \(\mathbf{4}\ \text{cm}\) |
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, \(2\sqrt{3} \times 3\sqrt{2} = 6\sqrt{6}\).
How do you divide surds?
Divide the numbers in front and divide the numbers under the roots, then simplify. For example, \(18\sqrt{40} \div 6\sqrt{5} = 3\sqrt{8} = 6\sqrt{2}\).
What is root 5 times root 5?
\(5\). Any square root multiplied by itself gives the number under the root.
How do you expand brackets with surds?
Multiply every term in the first bracket by every term in the second, then simplify each surd and collect like terms. Use the perfect square and difference of two squares patterns when they fit.
Can you multiply unlike surds?
Yes. Unlike surds cannot be added, but they can always be multiplied: \(\sqrt{2} \times \sqrt{3} = \sqrt{6}\).