Surds
Rationalising the denominator
In this note
- 1.18Rationalise a denominator \(\sqrt a\) or \(b\sqrt a\).
- 1.18Rationalise a denominator \(a\pm\sqrt b\) using the conjugate.
- 1.18Give the answer in its simplest form, e.g. \(a+b\sqrt c\).
Before you start: simplifying surds, especially \((a+\sqrt b)(a-\sqrt b)=a^{2}-b\).
Formula bank
Surd on the bottom
To rationalise the denominator is to make the bottom of a fraction a whole number. Multiply top AND bottom by the same thing (that is multiplying by 1), chosen so the surd on the bottom disappears.
General methodHow to rationalise \(\dfrac{k}{\sqrt a}\) or \(\dfrac{k}{b\sqrt a}\)
- Multiply top and bottom by \(\sqrt a\) (just the surd part).
- The bottom becomes \(a\) (or \(b\times a\)).
- Simplify: cancel common factors, simplify any surd on top.
Example Rationalise \(\dfrac6{\sqrt{10}}\).
× \(\dfrac{\sqrt{10}}{\sqrt{10}}\), then cancel.
- \(\dfrac{6\sqrt{10}}{10}\)
Answer \(\dfrac{3\sqrt{10}}5\)
Try it yourself
Show answer
Answer \(3\sqrt5\)
Show answer
Answer \(5\sqrt2\)
Show answer
Answer \(\dfrac{\sqrt{10}}4\)
A sum or difference on the bottom
General methodHow to rationalise \(\dfrac{k}{a\pm\sqrt b}\)
- Write the conjugate: change the sign in the middle (\(2-\sqrt5\) → \(2+\sqrt5\)).
- Multiply top and bottom by it.
- Bottom: \(a^{2}-b\) (no surd). Top: expand if there is a bracket.
- Divide every term on top by the bottom; simplify.
Multiplying \(\dfrac1{\sqrt7+1}\) by \(\sqrt7\) does NOT work: the bottom becomes \(7+\sqrt7\).
Example Rationalise and simplify \(\dfrac8{\sqrt3-1}\).
Multiply by \(\dfrac{\sqrt3+1}{\sqrt3+1}\).
- bottom: \(3-1=2\)
- \(\dfrac{8(\sqrt3+1)}{2}=4(\sqrt3+1)\)
Answer \(4\sqrt3+4\)
Try it yourself
Hint
Show answer
Answer \(6-2\sqrt2\)
Hint
Show answer
Answer \(-6-3\sqrt5\)
Show answer
Answer \(\sqrt3+\sqrt2\)
Ways they test it
Type 1 Single surd
"Rationalise the denominator: \(\dfrac{12}{\sqrt6}\)." (2 marks)
Spot it a surd alone on the bottom.
- × √a / √a.
- Simplify.
For example \(\dfrac{12}{\sqrt6}\)
- \(\dfrac{12\sqrt6}{6}\)
Answer \(2\sqrt6\)
Type 2 Conjugate
"Rationalise and simplify \(\dfrac8{\sqrt3-1}\)." (3 marks)
Spot it a sum or difference on the bottom.
- × conjugate (M1).
- Rational bottom (M1).
- Simplest form (A1).
Type 3 In the form a + √b
"Write \(\dfrac{11}{4-\sqrt5}\) in the form \(a+\sqrt b\)."
Spot it "in the form".
- Conjugate.
- Simplify to match the form.
For example \(\dfrac{11}{4-\sqrt5}\)
- \(\dfrac{11(4+\sqrt5)}{16-5}\)
Answer \(4+\sqrt5\)
Type 4 Show that (surds on top too)
"Show that \(\dfrac{2+\sqrt3}{2-\sqrt3}=7+4\sqrt3\)."
Spot it "show that".
- × conjugate.
- Expand the top carefully; every line written.
For example \(\dfrac{(2+\sqrt3)^{2}}{4-3}\)
- \(4+4\sqrt3+3\)
Answer \(7+4\sqrt3\)
Mistakes to avoid
- ✗ \(\dfrac1{\sqrt7+1}\times\dfrac{\sqrt7}{\sqrt7}\)✓ use the conjugateThe surd stays on the bottom.
- ✗ \((\sqrt3-1)(\sqrt3+1)=3+1\)✓ 3 − 1 = 2Difference of two squares.
- ✗ Multiplying only the bottom✓ top AND bottomOtherwise the value changes.
- ✗ \(\dfrac{20(4-\sqrt6)}{10}\) left✓ \(8-2\sqrt6\)Simplest form.
Summary
- Bottom \(\sqrt a\): × \(\dfrac{\sqrt a}{\sqrt a}\).
- Bottom \(a\pm\sqrt b\): × the conjugate over itself; the bottom becomes \(a^{2}-b\).
- Then simplify and cancel.
If you see …
- a surd alone on the bottom× that surd, top and bottom.
- a sum or difference on the bottomthe conjugate.
Practice
Practice: easy, medium, hard
Easy
Show working
- \(\dfrac{12\sqrt6}{6}\)
Answer \(2\sqrt6\)
Show working
- \(\dfrac{9\sqrt3}{3}\)
Answer \(3\sqrt3\)
Show answer
Answer \(3+\sqrt2\)
Medium
Guide me
What is it asking? a simplified fraction
What do we know? \(\sqrt3-1\) on the bottom
Plan conjugate, bottom 2
Show working
- \(\dfrac{8(\sqrt3+1)}{2}\)
Answer \(4+4\sqrt3\)
Guide me
What is it asking? a + √b
What do we know? \(4-\sqrt5\)
Plan conjugate, bottom 11
Show working
- \(\dfrac{11(4+\sqrt5)}{11}\)
Answer \(4+\sqrt5\)
Guide me
What is it asking? simplest form
What do we know? two surds on the bottom
Plan conjugate, bottom 6 − 2
Show working
- \(\dfrac{4(\sqrt6-\sqrt2)}{4}\)
Answer \(\sqrt6-\sqrt2\)
Hard
Guide me
What is it asking? simplest form
What do we know? \(\sqrt7-2\)
Plan conjugate; bottom 7 − 4 = 3
Show working
- \(\dfrac{6(\sqrt7+2)}{3}\)
Answer \(2\sqrt7+4\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| × \(\dfrac{\sqrt7+2}{\sqrt7+2}\) | M1 | ||
| denominator 3 | M1 | ||
| \(2\sqrt7+4\) | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? the denominator after × conjugate
What do we know? \((2-\sqrt3)(2+\sqrt3)\)
Plan 4 − 3
Show working
- \(\dfrac{(2+\sqrt3)^{2}}{4-3}=7+4\sqrt3\)
Answer 1
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| × conjugate | M1 | ||
| \((2+\sqrt3)^{2}=7+4\sqrt3\) | M1 | ||
| denominator 1, conclusion | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? a + b√5
What do we know? surds top and bottom
Plan conjugate; top squared; bottom 4
Show working
- \(\dfrac{(\sqrt5+1)^{2}}{4}=\dfrac{6+2\sqrt5}{4}\)
- \(=\tfrac32+\tfrac12\sqrt5\)
Answer \(\tfrac32+\tfrac12\sqrt5\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| × conjugate | M1 | ||
| \(\dfrac{6+2\sqrt5}{4}\) | M1 | ||
| \(\tfrac32+\tfrac12\sqrt5\) | A1 | ||
| Total: 3 marks | |||
Can I do this?
- I can rationalise a denominator √a or b√a.
- I can rationalise a denominator a ± √b with the conjugate.
- I can simplify the result, including forms like a + b√c.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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