CIE IGCSE Mathematics (0580) Surds Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Surds Study Notes
LEARNING OBJECTIVE
- Understanding Surd
Key Concepts:
- Surd
- Rationalising denominators
Surd
Surd
A surd is an irrational number that cannot be simplified to remove the square root (or cube root, etc.) and still remain exact. It is usually left in root form (not as a decimal).
- \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \) are examples of surds — they are irrational and non-terminating.
- \( \sqrt{4} = 2 \) is not a surd because it simplifies to a rational number.
Surds are used to give exact answers without rounding. They maintain accuracy in geometry (e.g. Pythagoras’ Theorem) and algebra.
Rules of Surds:
- \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \)
- \( \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \)
- \( (\sqrt{a})^2 = a \)
- \( \sqrt{a^2} = a \), if \( a \geq 0 \)
Simplifying Surds:
Break down the number inside the square root into factors, one of which is a perfect square.
- \( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \)
- \( \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \)
Adding & Subtracting Surds:
You can only add or subtract surds if they have the same radical part (just like like terms in algebra).
- \( 3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2} \)
- \( 4\sqrt{3} – 2\sqrt{3} = 2\sqrt{3} \)
- \( 2\sqrt{5} + 3\sqrt{2} \) cannot be simplified further
Multiplying & Dividing Surds:
- \( \sqrt{3} \times \sqrt{12} = \sqrt{36} = 6 \)
- \( \frac{\sqrt{18}}{\sqrt{2}} = \sqrt{9} = 3 \)
Example :
Simplify \( \sqrt{75} \)
▶️ Answer/Explanation
\( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \)
Example :
Simplify \( 2\sqrt{7} + 5\sqrt{7} – \sqrt{7} \)
▶️ Answer/Explanation
\( (2 + 5 – 1)\sqrt{7} = 6\sqrt{7} \)
Example :
Simplify \( \sqrt{8} \times \sqrt{2} \)
▶️ Answer/Explanation
\( \sqrt{8 \times 2} = \sqrt{16} = 4 \)
Rationalise the Denominator
Rationalise the Denominator
Rationalising the denominator means rewriting a fraction so that the denominator no longer contains a surd (a square root). This is done for clarity and standard form in mathematics.
It is not considered proper to leave a square root in the denominator. Rationalising helps simplify the expression and prepare it for further operations.
Rule:
- Multiply the numerator and denominator by the surd in the denominator.
- This uses the identity: \( \frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a} \)
Example:
Rationalise: \( \frac{1}{\sqrt{5}} \)
▶️ Answer/Explanation
Multiply numerator and denominator by \( \sqrt{5} \):
\( \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5} \)
Example:
Rationalise: \( \frac{3}{2\sqrt{7}} \)
▶️ Answer/Explanation
Multiply numerator and denominator by \( \sqrt{7} \):
\( \frac{3}{2\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{2 \times 7} = \frac{3\sqrt{7}}{14} \)
Example:
Rationalise: \( \frac{4}{3 + \sqrt{2}} \)
▶️ Answer/Explanation
This has a binomial denominator. Multiply by the conjugate \( (3 – \sqrt{2}) \):
$\frac{4}{3 + \sqrt{2}} \times \frac{3 – \sqrt{2}}{3 – \sqrt{2}} = \frac{4(3 – \sqrt{2})}{(3 + \sqrt{2})(3 – \sqrt{2})} $
Denominator: \( 3^2 – (\sqrt{2})^2 = 9 – 2 = 7 \)
Numerator: \( 4(3 – \sqrt{2}) = 12 – 4\sqrt{2} \)
Final Answer: \( \frac{12 – 4\sqrt{2}}{7} \)