Surds
Understand surds for Victorian Year 11 Mathematical Methods (VCAA) — exact numbers written with a square root sign, such as root two, that cannot be written as a simple fraction.
You will learn to simplify surds using perfect-square factors, add, subtract and multiply them, and rationalise the denominator of a fraction, giving the exact answers used throughout algebra and geometry.
Theory
In Year 11 Mathematical Methods (Unit 1), a surd is an irrational number written with a root sign, such as \(\sqrt{2}\) or \(\sqrt{5}\). This one subtopic covers both simplifying surds — writing a root in simplest exact form by extracting perfect-square factors and doing surd arithmetic — and rationalising the denominator, rewriting a fraction so no surd is left on the bottom using \(\dfrac{\sqrt{b}}{\sqrt{b}}\) or a conjugate.
A surd is a square root of a natural number that is irrational — it cannot be written as an exact fraction or decimal. \(\sqrt{2},\ \sqrt{3}\) and \(\sqrt{5}\) are surds, but \(\sqrt{9}=3\) and \(\sqrt{16}=4\) are not, because \(9\) and \(16\) are perfect squares.
A surd is in simplest form when the number under the root has no perfect-square factor larger than \(1\). We simplify by finding the largest perfect-square factor and splitting the root with \(\sqrt{ab}=\sqrt{a}\,\sqrt{b}\).
Like surds have the same number under the root (for example \(3\sqrt{2}\) and \(5\sqrt{2}\)). Only like surds can be added or subtracted, by collecting their coefficients.
To rationalise the denominator is to remove the surd from the bottom of a fraction while keeping the fraction equal in value, by multiplying by a cleverly chosen form of \(1\). For a single surd denominator, multiply top and bottom by that surd, since \(\sqrt{b}\times\sqrt{b}=b\) is rational. For a binomial surd denominator such as \(a+\sqrt{b}\), multiply by its conjugate \(a-\sqrt{b}\); by the difference of two squares \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\), which is rational.
The product and quotient rules for surds (for \(a,b\ge 0\)):
Collecting like surds (same number under the root):
Single-surd denominator — multiply by \(\dfrac{\sqrt{b}}{\sqrt{b}}\):
Binomial denominator — multiply by the conjugate:
How to simplify a surd
- Factor: find the largest perfect-square factor of the number under the root (\(4,9,16,25,\dots\)).
- Split: use \(\sqrt{ab}=\sqrt{a}\,\sqrt{b}\) and take the exact root of the perfect square.
- Combine: for sums, collect like surds; for products, multiply coefficients and roots (\(p\sqrt{a}\times q\sqrt{b}=pq\sqrt{ab}\)), then re-simplify.
How to rationalise a denominator
- Choose the multiplier: a single surd \(\sqrt{b}\) needs \(\dfrac{\sqrt{b}}{\sqrt{b}}\); a binomial \(a\pm\sqrt{b}\) needs its conjugate \(a\mp\sqrt{b}\).
- Multiply: multiply numerator and denominator by that multiplier, expanding with \(\sqrt{b}\,\sqrt{b}=b\) or \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\).
- Simplify: simplify any surd in the numerator and cancel common factors for the simplest exact form.
Largest perfect-square factor — \(72=36\times 2\):
| \(\sqrt{72}\) | \(=\) | \(\sqrt{36\times 2}\) |
Split the root and take \(\sqrt{36}\):
| \(=\) | \(\sqrt{36}\,\sqrt{2}\) | |
| \(=\) | \(6\sqrt{2}\) |
\(\sqrt{72}=6\sqrt{2}\).
Simplify each surd first:
| \(\sqrt{50}\) | \(=\) | \(\sqrt{25\times 2}=5\sqrt{2}\) |
| \(\sqrt{18}\) | \(=\) | \(\sqrt{9\times 2}=3\sqrt{2}\) |
| \(\sqrt{8}\) | \(=\) | \(\sqrt{4\times 2}=2\sqrt{2}\) |
They are all like surds — collect the coefficients:
| \(5\sqrt{2}-3\sqrt{2}+2\sqrt{2}\) | \(=\) | \((5-3+2)\sqrt{2}\) |
| \(=\) | \(4\sqrt{2}\) |
\(\sqrt{50}-\sqrt{18}+\sqrt{8}=4\sqrt{2}\).
Multiply coefficients, and the numbers under the roots:
| \(2\sqrt{6}\times 5\sqrt{3}\) | \(=\) | \((2\times 5)\sqrt{6\times 3}\) |
| \(=\) | \(10\sqrt{18}\) |
Simplify the surd — \(18=9\times 2\):
| \(10\sqrt{18}\) | \(=\) | \(10\sqrt{9}\,\sqrt{2}\) |
| \(=\) | \(10\times 3\sqrt{2}\) | |
| \(=\) | \(30\sqrt{2}\) |
\(2\sqrt{6}\times 5\sqrt{3}=30\sqrt{2}\).
Multiply top and bottom by \(\sqrt{3}\):
| \(\dfrac{6}{\sqrt{3}}\) | \(=\) | \(\dfrac{6}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}\) |
| \(=\) | \(\dfrac{6\sqrt{3}}{3}\) |
Simplify the fraction \(\dfrac{6}{3}=2\):
| \(\dfrac{6\sqrt{3}}{3}\) | \(=\) | \(2\sqrt{3}\) |
\(\dfrac{6}{\sqrt{3}}=2\sqrt{3}\).
Multiply by the conjugate \(3-\sqrt{5}\):
| \(\dfrac{4}{3+\sqrt{5}}\) | \(=\) | \(\dfrac{4}{3+\sqrt{5}}\times\dfrac{3-\sqrt{5}}{3-\sqrt{5}}\) |
Denominator is a difference of squares:
| \((3+\sqrt{5})(3-\sqrt{5})\) | \(=\) | \(3^{2}-(\sqrt{5})^{2}\) |
| \(=\) | \(9-5=4\) |
So the fraction becomes:
| \(=\) | \(\dfrac{4(3-\sqrt{5})}{4}\) | |
| \(=\) | \(3-\sqrt{5}\) |
\(\dfrac{4}{3+\sqrt{5}}=3-\sqrt{5}\).
Multiply by the conjugate \(2+\sqrt{3}\):
| \(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\) | \(=\) | \(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\times\dfrac{2+\sqrt{3}}{2+\sqrt{3}}\) |
Denominator (difference of squares):
| \((2-\sqrt{3})(2+\sqrt{3})\) | \(=\) | \(2^{2}-(\sqrt{3})^{2}=4-3=1\) |
Numerator (expand the square):
| \((2+\sqrt{3})^{2}\) | \(=\) | \(4+2\cdot 2\sqrt{3}+3\) |
| \(=\) | \(7+4\sqrt{3}\) |
Divide by \(1\):
| \(=\) | \(\dfrac{7+4\sqrt{3}}{1}=7+4\sqrt{3}\) |
\(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=7+4\sqrt{3}\).
Common pitfalls
Frequently asked questions
What is a surd?
A surd is the square root of a natural number that is irrational, such as \(\sqrt{2}\) or \(\sqrt{7}\). Roots of perfect squares like \(\sqrt{9}=3\) are not surds.
How do you simplify a surd like \(\sqrt{72}\)?
Find the largest perfect-square factor: \(72=36\times 2\). Then \(\sqrt{72}=\sqrt{36}\,\sqrt{2}=6\sqrt{2}\).
When can you add or subtract surds?
Only when they are like surds — the same number under the root. Simplify each surd first, then collect coefficients, e.g. \(5\sqrt{2}+2\sqrt{2}=7\sqrt{2}\).
How do you multiply two surds?
Multiply the coefficients and multiply the numbers under the roots: \(p\sqrt{a}\times q\sqrt{b}=pq\sqrt{ab}\), then simplify the result.
What does rationalising the denominator mean?
It means rewriting a fraction so there is no surd in the denominator, without changing its value, by multiplying the top and bottom by a suitable surd or conjugate.
How do you rationalise \(\dfrac{a}{\sqrt{b}}\)?
Multiply numerator and denominator by \(\sqrt{b}\): \(\dfrac{a}{\sqrt{b}}\times\dfrac{\sqrt{b}}{\sqrt{b}}=\dfrac{a\sqrt{b}}{b}\), then simplify.
What is a conjugate and when do you use it?
The conjugate of \(a+\sqrt{b}\) is \(a-\sqrt{b}\). Use it when the denominator is a binomial surd, because \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\) is rational.
Is \(\sqrt{a+b}\) the same as \(\sqrt{a}+\sqrt{b}\)?
No. The root does not split over a sum. For instance \(\sqrt{9+16}=\sqrt{25}=5\), but \(\sqrt{9}+\sqrt{16}=3+4=7\).