A conjugate turns a surd denominator into a whole number, because the difference of two squares makes the surd disappear. That is the entire trick.
Today's hook, Before calculators, dividing by $1.732...$ by hand was painful, while dividing by a whole number was not. Rationalising the denominator was invented for that, and it survives because it gives one agreed exact form for an answer.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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Multiply $(2 + \sqrt{3})(2 - \sqrt{3})$. What happens to the surd, and why? Predict what $(5 + \sqrt{7})(5 - \sqrt{7})$ will give before you work it out.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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The conjugate kills the surd
Work through the core explanation before applying it.
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The conjugate kills the surd
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The conjugate of a binomial surd is the same expression with the middle sign reversed. Multiplying a binomial surd by its conjugate always gives a rational number, because it is a difference of two squares.
$(2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - \left(\sqrt{3}\right)^2 = 4 - 3 = 1$. The cross terms $-2\sqrt{3}$ and $+2\sqrt{3}$ cancel, and squaring the surd removes the root.
The conjugate of $3 - 2\sqrt{5}$ is $3 + 2\sqrt{5}$. The first term keeps its sign.
Multiply top and bottom
You are multiplying by a form of 1, so the value does not change. Multiplying only the denominator changes the answer.
One surd needs only itself
For $\frac{6}{\sqrt{3}}$ there is no binomial, so multiply top and bottom by $\sqrt{3}$, not by a conjugate.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$, and of $a - \sqrt{b}$ is $a + \sqrt{b}$.
$\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right) = a^2 - b$, which is always rational.
For a single surd denominator, multiply the numerator and denominator by that surd.
For a binomial surd denominator, multiply the numerator and denominator by the conjugate.
Understand
Concepts
Why multiplying by the conjugate removes the surd, via the difference of two squares.
Why you must multiply the numerator as well as the denominator.
Why rationalised form is the conventional way to present an exact answer.
Can do
Skills
Write down the conjugate of any binomial surd.
Rationalise a denominator containing a single surd.
Rationalise a denominator containing a binomial surd, and simplify the result.
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Key terms
Rationalise the denominatorRewrite a fraction so no surd is left underneath, without changing its value. Like this: $\frac{6}{\sqrt{3}}$ becomes $2\sqrt{3}$.
ConjugateThe same binomial surd with the middle sign reversed, used to clear a surd from a denominator. Like this: the conjugate of $2 + \sqrt{3}$ is $2 - \sqrt{3}$.
Binomial surdA two-term expression where at least one term is a surd. Like this: $5 - \sqrt{7}$ and $\sqrt{2} + \sqrt{3}$ are binomial surds.
Difference of two squaresThe pattern $(a+b)(a-b) = a^2 - b^2$, which is why a conjugate works. Like this: $(4+\sqrt{5})(4-\sqrt{5}) = 16 - 5 = 11$.
Multiplying by 1Multiplying a fraction by something over itself, which changes its appearance but not its value. Like this: $\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$ is multiplying by 1.
Exact formAn answer left in surd form rather than rounded to a decimal. Like this: $2\sqrt{3}$ is exact, while $3.46$ is a rounded approximation.
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Rationalising a single surd denominator
Work through the core explanation before applying it.
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Rationalising a single surd denominator
core concept
When the denominator is a single surd, multiply the numerator and the denominator by that surd. Because $\sqrt{a}\sqrt{a} = a$, the denominator becomes rational.
For $\dfrac{6}{\sqrt{3}}$, multiply top and bottom by $\sqrt{3}$: $\dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$.
You are multiplying by $\dfrac{\sqrt{3}}{\sqrt{3}}$, which equals 1, so the value is unchanged. Only the appearance changes.
Simplify at the end. $\dfrac{6\sqrt{3}}{3}$ is correct but not finished. Cancel the 6 and the 3 to get $2\sqrt{3}$.
Quick check: rationalise $\dfrac{10}{\sqrt{5}}$.
For a single surd denominator, multiply the numerator and denominator by that surd. $\sqrt{a}\sqrt{a} = a$ clears the root. You are multiplying by 1, so the value is unchanged. Always cancel at the end.
Pause, copy the single-surd method with the worked $\frac{6}{\sqrt{3}} = 2\sqrt{3}$, and note that multiplying top and bottom by the same thing is multiplying by 1, into your book.
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The conjugate, and why it works
core concept
We just saw that a single surd denominator clears by multiplying by that surd. That raises a question: what do you do when the denominator has two terms, such as $2 + \sqrt{3}$, where multiplying by $\sqrt{3}$ leaves a surd behind? This card answers it → multiply by the conjugate, which uses the difference of two squares to remove the surd completely.
The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$: same terms, middle sign reversed. Only the sign between the terms changes.
Multiplying a binomial surd by its conjugate gives a difference of two squares: $\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right) = a^2 - b$. The cross terms cancel and the surd is squared away.
For example $(4 + \sqrt{5})(4 - \sqrt{5}) = 16 - 5 = 11$, a whole number. That is exactly what a denominator needs.
Why multiplying by $\sqrt{3}$ is not enough. $(2 + \sqrt{3})\sqrt{3} = 2\sqrt{3} + 3$, which still contains a surd. Only the conjugate clears it.
The conjugate of $3 - 2\sqrt{5}$ is $3 + 2\sqrt{5}$. Which expression is NOT another way of writing that conjugate?
The conjugate reverses the middle sign only. Multiplying a binomial surd by its conjugate gives $a^2 - b$, always rational, because the cross terms cancel and the surd gets squared.
Pause, copy the definition of a conjugate, the identity $\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right) = a^2 - b$, and the worked $(4+\sqrt{5})(4-\sqrt{5}) = 11$, into your book.
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Rationalising a binomial surd denominator
core concept
We just saw that a conjugate turns a binomial surd into a rational number. That raises a question: how do you use that on an actual fraction without changing its value? This card answers it → multiply the numerator and denominator by the conjugate, then expand and simplify.
Multiply the numerator and the denominator by the conjugate of the denominator. For $\dfrac{4}{2 + \sqrt{3}}$, that conjugate is $2 - \sqrt{3}$.
The denominator becomes $(2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1$, and the numerator becomes $4(2 - \sqrt{3}) = 8 - 4\sqrt{3}$.
So $\dfrac{4}{2+\sqrt{3}} = 8 - 4\sqrt{3}$. Expand the numerator fully and simplify or cancel where possible.
Do not expand the denominator the long way. Use $a^2 - b$ directly. Expanding all four terms wastes time and invites a sign error.
Fill the blank: to rationalise $\dfrac{1}{5 - \sqrt{2}}$, multiply top and bottom by $5 + \sqrt{\;}$ where the number under the root is .
Multiply the numerator and denominator by the conjugate of the denominator. Use $a^2 - b$ for the new denominator rather than expanding four terms, expand the numerator fully, then simplify or cancel.
Pause, copy the binomial method with the worked $\frac{4}{2+\sqrt{3}} = 8 - 4\sqrt{3}$, and the shortcut that the new denominator is $a^2 - b$, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · SINGLE SURD DENOMINATOR
Rationalise the denominator of $\dfrac{6}{\sqrt{3}}$.
Difference of two squares, both surds squared away.
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Numerator: $\sqrt{2}\left(\sqrt{5}+\sqrt{2}\right) = \sqrt{10} + 2$, so the answer is $\dfrac{\sqrt{10} + 2}{3}$
Expand, using $\sqrt{2}\sqrt{2} = 2$.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Rationalise $\dfrac{8}{\sqrt{2}}$.
Write down the conjugate of $7 + \sqrt{5}$.
Evaluate $(3 - \sqrt{2})(3 + \sqrt{2})$.
Rationalise $\dfrac{1}{4 - \sqrt{3}}$.
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Revisit your prediction
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Revisit your prediction
You predicted the value of $(5 + \sqrt{7})(5 - \sqrt{7})$ at the start. Confirm it now, and explain in one sentence why the answer is always a whole number when $a$ and $b$ are whole numbers.
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Multiple choice
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Multiple choice
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. Rationalise the denominator of $\dfrac{5}{3 - \sqrt{2}}$, leaving your answer in exact form. (3 marks)
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ApplyBand 43 marks
Q2. Simplify $\dfrac{\sqrt{3}}{\sqrt{7} + \sqrt{3}}$ by rationalising the denominator. (3 marks)
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UnderstandBand 32 marks
Q3. Explain why multiplying $\dfrac{1}{2 + \sqrt{5}}$ by $\dfrac{\sqrt{5}}{\sqrt{5}}$ does not rationalise the denominator, but multiplying by the conjugate does. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $4\sqrt{2}$. Practice 2: $7 - \sqrt{5}$. Practice 3: $9 - 2 = 7$. Practice 4: $\dfrac{4+\sqrt{3}}{13}$.
Q1 (3 marks): Multiply top and bottom by $3 + \sqrt{2}$ [1]. Denominator $= 9 - 2 = 7$ [1]. Answer $= \dfrac{5(3+\sqrt{2})}{7} = \dfrac{15 + 5\sqrt{2}}{7}$ [1].
Q2 (3 marks): Multiply top and bottom by $\sqrt{7} - \sqrt{3}$ [1]. Denominator $= 7 - 3 = 4$ [1]. Numerator $= \sqrt{3}\left(\sqrt{7}-\sqrt{3}\right) = \sqrt{21} - 3$, so the answer is $\dfrac{\sqrt{21} - 3}{4}$ [1].
Q3 (2 marks): $(2+\sqrt{5})\sqrt{5} = 2\sqrt{5} + 5$, which still contains a surd, so the denominator is not rational [1]. The conjugate $2 - \sqrt{5}$ gives $4 - 5 = -1$, a rational number, because the cross terms cancel and the surd is squared [1].
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Review and finish
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Boss battle · Conjugate Clash
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Clear surds from denominators at speed, choosing between a single surd and a conjugate. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.