Year 10 Mathematics · Number and Algebra · 20m read

Surds and Exact Values

Ever wondered why mathematicians refuse to round 2 to 3.14? Surds are numbers left in square root form because they're infinitely precise—no rounding errors, no approximations. You'll learn how to simplify surds, add and multiply them, and keep your answers exact, which matters everywhere from VCE exams to engineering where precision counts.

Learning objective: Simplify and perform operations with surds, leaving answers in exact form.

By the end of this lesson you can:

I simplify surds.
I add, subtract, multiply, and divide surds.
I rationalise the denominator.

Surds are exact bullseyes. √2 ≈ 1.414 is close, but √2 hits the exact centre. Keeping surds avoids rounding errors.

Surds and exact arithmetic

A SURD is an irrational root that cannot be simplified to a rational number.
$\sqrt{2}, \sqrt{3}, \sqrt{7}, 2\sqrt{5}$ are surds. $\sqrt{4} = 2$ is NOT a surd.

### Simplifying Surds
Find the largest perfect-square factor.
$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$
$\sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$

### Adding and Subtracting Surds
Only LIKE surds (same radicand) can be combined:
$3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$
$3\sqrt{3} - \sqrt{3} = 2\sqrt{3}$
$2\sqrt{5} + 3\sqrt{2}$ cannot be simplified (unlike surds).

Simplify first, then combine:
$\sqrt{75} + \sqrt{48} = 5\sqrt{3} + 4\sqrt{3} = 9\sqrt{3}$

### Multiplying Surds
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
$3\sqrt{2} \times 4\sqrt{5} = 12\sqrt{10}$
$(2\sqrt{3})^2 = 4 \times 3 = 12$
$(\sqrt{a})^2 = a$

Expanding with surds — use FOIL:
$(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$ (difference of two squares!)
$(1 + \sqrt{5})^2 = 1 + 2\sqrt{5} + 5 = 6 + 2\sqrt{5}$

### Dividing Surds
$\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$

$\dfrac{6\sqrt{10}}{3\sqrt{2}} = 2\sqrt{5}$

### Rationalising the Denominator
REMOVE THE SURD from the denominator.

Type 1 — single surd denominator:
$\dfrac{3}{\sqrt{2}} = \dfrac{3}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{3\sqrt{2}}{2}$

Type 2 — binomial denominator: multiply by the CONJUGATE.
$\dfrac{5}{3 + \sqrt{2}} = \dfrac{5(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} = \dfrac{5(3 - \sqrt{2})}{9 - 2} = \dfrac{5(3 - \sqrt{2})}{7}$

The conjugate of $a + b$ is $a - b$. Their product $= a^2 - b^2$ (DOTS).

### Exact Trig Values

| Angle | $\sin$ | $\cos$ | $\tan$ |
|-------|--------|--------|--------|
| 30° | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ |
| 45° | $\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | $1$ |
| 60° | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |

Memorise these! They appear constantly in VCE.

Surd Rules

Product rule: \sqrt{a} \times \sqrt{b} = \sqrt{ab}
Quotient rule: \dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}
Square of surd: (\sqrt{a})^2 = a
Conjugate product (DOTS): (a + \sqrt{b})(a - \sqrt{b}) = a^2 - b

Surd checklist

Like surds only: Only combine $n\sqrt{a} + m\sqrt{a}$. $\sqrt{2} + \sqrt{3}$ CANNOT be simplified.
Simplify first: Always simplify each surd before combining.
Rationalise always: Final answer must not have a surd in the denominator (unless told otherwise).

Worked example (Medium)

Simplify $\sqrt{75} - 2\sqrt{3} + \sqrt{12}$.

Simplify each. $\sqrt{75} = 5\sqrt{3}$; $\sqrt{12} = 2\sqrt{3}$.
Collect like surds. $5\sqrt{3} - 2\sqrt{3} + 2\sqrt{3} = 5\sqrt{3}$.

Answer: $5\sqrt{3}$.

Worked example (Hard)

Rationalise and simplify $\dfrac{4 + \sqrt{3}}{2 - \sqrt{3}}$.

Conjugate. Multiply by $\dfrac{2 + \sqrt{3}}{2 + \sqrt{3}}$.
Denominator. $(2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1$.
Numerator. $(4 + \sqrt{3})(2 + \sqrt{3}) = 8 + 4\sqrt{3} + 2\sqrt{3} + 3 = 11 + 6\sqrt{3}$.
Result. $\dfrac{11 + 6\sqrt{3}}{1} = 11 + 6\sqrt{3}$.

Answer: $11 + 6\sqrt{3}$.

Common mistakes

Writing $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ This is NOT valid. $\sqrt{9 + 16} = \sqrt{25} = 5 \neq \sqrt{9} + \sqrt{16} = 3 + 4 = 7$. Fix: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. ONLY product and quotient can be split: $\sqrt{ab} = \sqrt{a}\sqrt{b}$.
Not simplifying surds fully — stopping at a non-perfect-square factor $\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}$ is not fully simplified because $18 = 9 \times 2$ and 9 is a perfect square. Fix: Find the LARGEST perfect square factor: $72 = 36 \times 2$, so $\sqrt{72} = 6\sqrt{2}$.
Forgetting the conjugate when rationalising a binomial denominator Multiplying by $\frac{1}{3 + \sqrt{2}}$ or by $\frac{\sqrt{2}}{\sqrt{2}}$ does not eliminate the surd from $3 + \sqrt{2}$. Fix: Multiply by the conjugate $\frac{3 - \sqrt{2}}{3 - \sqrt{2}}$. The product of conjugates is rational: $(3+\sqrt{2})(3-\sqrt{2}) = 9-2=7$.

Practice questions

$\sqrt{50}$ simplified is:
1. $5\sqrt{2}$
2. $25\sqrt{2}$
3. $2\sqrt{5}$
4. $10\sqrt{5}$
(1 marks)
Simplify $3\sqrt{8} + 2\sqrt{18} - \sqrt{2}$.
(2 marks)
Expand $(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})$.
(2 marks)
Rationalise the denominator of $\dfrac{2}{\sqrt{3} - 1}$.
(3 marks)

Going Deeper — Fractional Indices and Surds

Surds and fractional indices are closely related:
$$a^{1/2} = \sqrt{a}, \quad a^{1/3} = \sqrt[3]{a}, \quad a^{m/n} = \sqrt[n]{a^m}$$

This unifies index laws and surds into one framework.

Examples:
$8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$
$27^{-1/3} = \dfrac{1}{\sqrt[3]{27}} = \dfrac{1}{3}$
$16^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8$

Simplifying with fractional indices:
$\dfrac{a^{3/2}}{a^{1/2}} = a^{3/2 - 1/2} = a^1 = a$

This is the preferred form in VCE — expressions like $\sqrt[3]{a^2}$ are written as $a^{2/3}$.

Surds also appear in exact trigonometric values used throughout VCE Specialist Mathematics and Mathematical Methods. Being fluent with surd arithmetic is essential preparation.

Summary

Simplify by finding the largest perfect-square factor: $\sqrt{72} = 6\sqrt{2}$.
Add/subtract LIKE surds only (same radicand).
Multiply: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$; $(\sqrt{a})^2 = a$.
Rationalise type 1 (single surd): multiply by $\sqrt{a}/\sqrt{a}$.
Rationalise type 2 (binomial): multiply by conjugate $(a - \sqrt{b})/(a - \sqrt{b})$.