Rationalising with a Conjugate

Statement

If the denominator is of the form \(a + \sqrt{b}\), we rationalise by multiplying top and bottom by the conjugate \(a - \sqrt{b}\). This removes the surd from the denominator and leaves a simplified expression:

\[ \frac{1}{a + \sqrt{b}} \times \frac{a - \sqrt{b}}{a - \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}. \]

Why it’s true

• The product of conjugates uses the difference of two squares: \((a+\sqrt{b})(a-\sqrt{b}) = a^2 - b\).
• This eliminates the square root from the denominator, leaving a rational number below.

Recipe (how to use it)

1. Identify a denominator of the form \(a + \sqrt{b}\) or \(a - \sqrt{b}\).
2. Multiply numerator and denominator by the conjugate (\(a - \sqrt{b}\) or \(a + \sqrt{b}\)).
3. Simplify: denominator becomes \(a^2 - b\).

Spotting it

This appears when fractions have denominators that are part rational, part surd.

Common pairings

• Simplifying expressions with surds in denominators.
• Manipulating exact trigonometric values.
• Preparation for algebraic proof or calculus simplifications.

Mini examples

1. \(\tfrac{1}{2+\sqrt{3}} = \tfrac{2-\sqrt{3}}{4-3} = 2-\sqrt{3}\).
2. \(\tfrac{1}{3+\sqrt{2}} = \tfrac{3-\sqrt{2}}{9-2} = (3-\sqrt{2})/7\).

Pitfalls

• Forgetting to use the conjugate — multiplying by \(a+\sqrt{b}\) again won’t remove the surd.
• Incorrectly expanding the denominator — must use difference of squares.
• Not simplifying numerator fully when possible.

Exam strategy

• Always check if the denominator has a rational and a surd part.
• Show multiplication by the conjugate clearly to get full marks.
• Simplify the denominator to \(a^2 - b\), not leaving it expanded with surds.

Summary

When the denominator has both a number and a square root, rationalisation requires multiplying by the conjugate. This removes surds from the denominator and leaves an exact simplified fraction, which is standard GCSE practice.