Surds

Introduction to Surds

In your math journey so far, you’ve worked with square roots like \(\sqrt{25}\), which is simply \(5\). But what happens when you have a root that doesn't result in a whole number, like \(\sqrt{2}\)? If you put \(\sqrt{2}\) into a calculator, you get \(1.41421356...\) – a decimal that goes on forever without repeating.

A surd is an irrational number that is left in its root form (like \(\sqrt{3}\) or \(\sqrt{5}\)) to keep it exact. In your IGCSE Extended papers, you will often be asked to give an "exact answer," which usually means leaving your answer as a surd rather than a rounded decimal.

Note: This topic is for Extended candidates only.

The Golden Rule: Square Numbers

To master surds, you must be best friends with your square numbers. You will use these to "break down" larger surds. Keep this list handy:

\(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144...\)

1. Simplifying Surds

Simplifying a surd means making the number under the root sign as small as possible. We do this by finding the largest square number that divides into it.

The Rule: \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\)

Step-by-Step Example: Simplify \(\sqrt{50}\)

1. Find factors of \(50\). We look for a square number: \(25 \times 2 = 50\).
2. Write it as: \(\sqrt{25 \times 2}\).
3. Split them up: \(\sqrt{25} \times \sqrt{2}\).
4. Calculate the square root of the square number: \(5 \times \sqrt{2}\).
5. Final Answer: \(5\sqrt{2}\).

Common Mistake: Don't stop too early! If you simplified \(\sqrt{48}\) as \(\sqrt{4} \times \sqrt{12} = 2\sqrt{12}\), you aren't finished because \(12\) still has a square factor (\(4\)). Always look for the biggest square factor first (\(16 \times 3\)) to get \(4\sqrt{3}\) immediately.

2. Adding and Subtracting Surds

You can only add or subtract surds if they are "like terms" (the same number under the root sign). Think of it like Algebra!

Example:
\(3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}\) (Just like \(3x + 2x = 5x\))
\(7\sqrt{2} - \sqrt{2} = 6\sqrt{2}\)

Important: \(\sqrt{9} + \sqrt{16}\) is NOT \(\sqrt{25}\).
Actually, \(\sqrt{9} + \sqrt{16} = 3 + 4 = 7\), whereas \(\sqrt{25} = 5\). You cannot combine different roots by adding the numbers inside!

3. Multiplying and Dividing Surds

Multiplication and division are much more flexible than addition. You can combine any two surds into one root.

Rules:
1. \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\)
2. \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\)

Examples:
\(\sqrt{3} \times \sqrt{5} = \sqrt{15}\)
\(2\sqrt{3} \times 4\sqrt{2} = 8\sqrt{6}\) (Multiply the outside numbers together, then multiply the roots together).
\(\frac{\sqrt{20}}{\sqrt{5}} = \sqrt{4} = 2\)

4. Rationalising the Denominator

In mathematics, it is considered "untidy" to leave a square root on the bottom of a fraction (the denominator). Rationalising is the process of moving the root to the top.

Type 1: Simple Denominator
To get rid of \(\sqrt{a}\) on the bottom, multiply both the top and the bottom of the fraction by \(\sqrt{a}\). Remember that \(\sqrt{a} \times \sqrt{a} = a\).

Example: Rationalise \(\frac{3}{\sqrt{5}}\)
1. Multiply top and bottom by \(\sqrt{5}\): \(\frac{3 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}\)
2. This becomes: \(\frac{3\sqrt{5}}{5}\).
The root is now gone from the bottom! Success!

Type 2: Conjugate Pairs (Advanced)
If the denominator is something like \(2 + \sqrt{3}\), you multiply by its "opposite" (the conjugate): \(2 - \sqrt{3}\). This uses the difference of two squares to cancel out the root.

Quick Review: Surd Rules

• Simplify: Look for square number factors.
• Add/Subtract: Only if the roots are the same.
• Multiply/Divide: Combine them under a single root.
• Rationalise: Multiply the top and bottom by the root you want to remove.

Key Takeaway for the Exam

Surds are most likely to appear on Paper 2 (Non-Calculator). You will be expected to simplify your final answers fully. If you see \(\sqrt{8}\) in your working, always change it to \(2\sqrt{2}\). Practice your times tables and square numbers to make finding factors quick and easy!

Did you know? The square root of any prime number is always a surd (irrational). This is why you see \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{7}\) so often in your textbook!