Surds are a special way to write down exact values for irrational numbers, like $\sqrt{2}$. Instead of rounding, we keep the number in its root form to maintain perfect accuracy, which is crucial in fields like engineering, architecture, and computer graphics.
What Exactly is a Surd?
Rational vs. Irrational Roots
A surd is a root (like a square root) that cannot be simplified to a whole number or fraction. Its decimal value goes on forever without repeating.
- $\sqrt{4} = 2$. This is a rational number, so it is not a surd.
- $\sqrt{2} \approx 1.414…$. This cannot be simplified, so it is a surd.
At GCSE, surds are mainly square roots of non-square numbers.
Working with Surds: Step-by-Step
1. Simplifying Surds
Make the number inside the root as small as possible by finding the largest square number that is a factor.
Example: Simplify $\sqrt{50}$
- Find the largest square factor of 50. It’s 25. So, $50 = 25 \times 2$.
- Split the surd: $\sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2}$.
- Simplify: $5 \times \sqrt{2}$.
Answer: $5\sqrt{2}$
2. Adding & Subtracting Surds
You can only add or subtract surds if the number inside the root is the same. Think of them like algebraic terms ($3x + 2x = 5x$).
Example: Calculate $\sqrt{8} + \sqrt{18}$
- The surds are different, so simplify them first.
- $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.
- $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$.
- Now add: $2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$.
Answer: $5\sqrt{2}$
3. Multiplying Surds
Multiply the numbers outside the surds together, and multiply the numbers inside together.
Example: Calculate $2\sqrt{5} \times 3\sqrt{2}$
- Multiply the outside numbers: $2 \times 3 = 6$.
- Multiply the inside numbers: $5 \times 2 = 10$.
Answer: $6\sqrt{10}$
4. Expanding Brackets
Use the same FOIL or distributive method as in algebra. A key shortcut is the difference of two squares: $(a + \sqrt{b})(a – \sqrt{b}) = a^2 – b$.
Example: Expand $(2 + \sqrt{3})(1 + \sqrt{3})$
- First: $2 \times 1 = 2$.
- Outer: $2 \times \sqrt{3} = 2\sqrt{3}$.
- Inner: $\sqrt{3} \times 1 = \sqrt{3}$.
- Last: $\sqrt{3} \times \sqrt{3} = 3$.
- Combine: $2 + 2\sqrt{3} + \sqrt{3} + 3 = 5 + 3\sqrt{3}$.
Answer: $5 + 3\sqrt{3}$
Rationalising the Denominator
This is the process of removing a surd from the bottom of a fraction. It’s a key skill for simplifying your final answers.
Method 1: Single Surd Denominator
If the denominator is $\sqrt{a}$, multiply the top and bottom of the fraction by $\sqrt{a}$.
Example: Rationalise $\frac{10}{\sqrt{5}}$
$\frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{\sqrt{25}} = \frac{10\sqrt{5}}{5}$
Simplify the fraction: $10 \div 5 = 2$.
Answer: $2\sqrt{5}$
Method 2: Binomial Denominator
If the denominator is like $a + \sqrt{b}$, multiply the top and bottom by its conjugate, which is $a – \sqrt{b}$.
Example: Rationalise $\frac{1}{3 – \sqrt{2}}$
Multiply by the conjugate $3 + \sqrt{2}$:
$\frac{1}{3 – \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{3 + \sqrt{2}}{(3 – \sqrt{2})(3 + \sqrt{2})}$
The denominator becomes $3^2 – (\sqrt{2})^2 = 9 – 2 = 7$.
Answer: $\frac{3 + \sqrt{2}}{7}$
Tutor Insights
🤔 Common Misunderstandings
- Adding surds incorrectly: Thinking $\sqrt{a} + \sqrt{b} = \sqrt{a+b}$. This is wrong! You can only add “like” surds, e.g., $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$.
- Not simplifying fully: Leaving an answer as $\sqrt{12}$ instead of $2\sqrt{3}$. Always check for square factors!
📝 Common Exam Mistakes
- Forgetting the conjugate: Not using the conjugate pair when rationalising a binomial denominator.
- Sign errors: Making mistakes with negative signs when expanding brackets.
- Incomplete simplification: Leaving an answer like $\frac{2\sqrt{2}}{4}$ instead of simplifying it further to $\frac{\sqrt{2}}{2}$.
Practice Questions
- Simplify: a) $\sqrt{20}$, b) $\sqrt{75}$.
- Calculate and simplify: $\sqrt{27} + \sqrt{12}$.
- Expand and simplify: $(4 + \sqrt{7})(2 – \sqrt{7})$.
- Rationalise the denominator of $\frac{1}{3 – \sqrt{2}}$.
- A rectangle has a length of $(3 + \sqrt{2})$ cm and a width of $(3 – \sqrt{2})$ cm. Find its exact area.
Show Answers
- a) $\sqrt{20} = \sqrt{4 \times 5} = \mathbf{2\sqrt{5}}$. b) $\sqrt{75} = \sqrt{25 \times 3} = \mathbf{5\sqrt{3}}$.
- Working: $\sqrt{27} = 3\sqrt{3}$, $\sqrt{12} = 2\sqrt{3}$. So $3\sqrt{3} + 2\sqrt{3} = \mathbf{5\sqrt{3}}$.
- Working: $8 – 4\sqrt{7} + 2\sqrt{7} – 7 = \mathbf{1 – 2\sqrt{7}}$.
- Working: $\frac{1}{3 – \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{3 + \sqrt{2}}{9 – 2} = \mathbf{\frac{3 + \sqrt{2}}{7}}$.
- Working: Area $= (3 + \sqrt{2})(3 – \sqrt{2}) = 3^2 – (\sqrt{2})^2 = 9 – 2 = 7$.
Answer: $\mathbf{7 \text{ cm}^2}$.
FAQs
Q: What’s the real difference between $\sqrt{4}$ and $\sqrt{5}$?
A: $\sqrt{4}$ is a rational number because it simplifies to a whole number, 2, so it’s not a surd. $\sqrt{5}$ is an irrational number because its decimal goes on forever without repeating. Therefore, $\sqrt{5}$ is a surd.
Q: Why do we have to “rationalise the denominator”?
A: It’s a mathematical convention that makes answers standard and easier to work with. Historically, it was very difficult to divide by a long decimal like $\sqrt{2}$ without a calculator. Converting $\frac{1}{\sqrt{2}}$ to $\frac{\sqrt{2}}{2}$ makes the calculation much simpler.
Q: Are surds always square roots?
A: At GCSE level, when we talk about surds, we almost always mean irrational square roots. Technically, a surd can be any root (like $\sqrt[3]{10}$) that is irrational, but for your exams, the focus is on square roots.
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