Rationalising 1/(√a + √b)

Statement

When a denominator contains a sum of surds such as \( \sqrt{a} + \sqrt{b} \), we can rationalise it by multiplying top and bottom by the conjugate \( \sqrt{a} - \sqrt{b} \). This removes the surds from the denominator:

\[ \frac{1}{\sqrt{a} + \sqrt{b}} \times \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} - \sqrt{b}}{a - b}. \]

Why it’s true

• Multiplying conjugates uses the difference of two squares: \((\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b\).
• This eliminates the surds from the denominator while leaving the fraction equivalent.

Recipe (how to use it)

1. Identify the denominator, e.g. \(\sqrt{a} + \sqrt{b}\).
2. Multiply numerator and denominator by the conjugate, \(\sqrt{a} - \sqrt{b}\).
3. Simplify using \(a - b\) in the denominator.

Spotting it

Use this whenever you see fractions with denominators of the form \( \sqrt{a} + \sqrt{b} \) or \( \sqrt{a} - \sqrt{b} \).

Common pairings

• Surds simplification in algebra questions.
• Exact trig values involving surds.
• Rationalising denominators before further manipulation.

Mini examples

1. \(\tfrac{1}{\sqrt{2} + \sqrt{3}} = \tfrac{\sqrt{2} - \sqrt{3}}{2 - 3} = \tfrac{\sqrt{3} - \sqrt{2}}{1}\).
2. \(\tfrac{1}{\sqrt{5} + \sqrt{2}} = \tfrac{\sqrt{5} - \sqrt{2}}{3}\).

Pitfalls

• Forgetting to multiply top and bottom by the conjugate.
• Expanding incorrectly (must use difference of two squares).
• Leaving the denominator unsimplified.

Exam strategy

• Always show the multiplication by the conjugate step clearly.
• Remember to simplify surds fully where possible.
• Check if the denominator simplifies to a small integer.

Summary

Rationalising denominators ensures fractions are written in their simplest form. By multiplying by the conjugate, surds disappear from the denominator, leaving a neater and exact expression.