Year 10 Core — Rationalising the Denominator (4D)

Today's lesson

This builds straight on top of 4C — Multiplying surds. We use the same multiplying rule, but for a specific purpose: getting rid of a surd in the denominator of a fraction.

Learning intentions

Understand that a surd multiplied by itself gives a whole number ($\sqrt{x}\times\sqrt{x}=x$)
Know that rationalising a denominator means turning an irrational denominator into a rational (whole-number) one
Be able to rationalise denominators of the form $\sqrt{y}$ and $b\sqrt{y}$

Part 1 — The key idea (warm up, ~5 min)

A surd like $\dfrac{1}{\sqrt{2}}$ is a bit awkward — it's hard to estimate, and traditionally we like rational (whole-number) denominators. The trick is to multiply the fraction by $1$ in disguise:

📺 Key idea: multiply the top and bottom by the surd in the denominator. That's the same as multiplying by $1$, so the value doesn't change — but the denominator becomes a whole number.

The rule

$\dfrac{x}{\sqrt{y}}\;=\;\dfrac{x}{\sqrt{y}}\times\dfrac{\sqrt{y}}{\sqrt{y}}\;=\;\dfrac{x\sqrt{y}}{y}$

The expression $\dfrac{\sqrt{y}}{\sqrt{y}}=1$, so multiplying by it doesn't change the value — it just rewrites the fraction in a tidier form.

Building understanding 1 — simplify each:

$\dfrac{\sqrt{6}}{\sqrt{6}}$
$\dfrac{2\sqrt{5}}{4\sqrt{5}}$
$-\dfrac{\sqrt{8}}{\sqrt{2}}$
$\dfrac{\sqrt{72}}{\sqrt{2}}$

a) $1$
b) $\dfrac{2}{4}=\dfrac{1}{2}$
c) $-\sqrt{4}=-2$
d) $\sqrt{36}=6$

Building understanding 2 — state the missing number:

$\sqrt{3}\times\underline{\phantom{xx}}=3$
$\sqrt{10}\times\sqrt{10}=\underline{\phantom{xx}}$
$2\sqrt{5}\times\underline{\phantom{xx}}=10$
$\underline{\phantom{xx}}\times 3\sqrt{7}=21$

a) $\sqrt{3}$
b) $10$
c) $\sqrt{5}$ (since $2\sqrt{5}\times\sqrt{5}=2\times 5=10$)
d) $\sqrt{7}$ (since $\sqrt{7}\times 3\sqrt{7}=3\times 7=21$)

Part 2 — Worked examples (~15 min)

EXAMPLE 9a

Rationalise the denominator: $\dfrac{2}{\sqrt{3}}$.

Multiply top and bottom by $\sqrt{3}$:

$\dfrac{2}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$

Now you try: $\dfrac{3}{\sqrt{2}}=\;?$ Answer: $\dfrac{3\sqrt{2}}{2}$

EXAMPLE 9b

Rationalise the denominator: $\dfrac{3\sqrt{2}}{\sqrt{5}}$.

Multiply top and bottom by $\sqrt{5}$:

$\dfrac{3\sqrt{2}}{\sqrt{5}}=\dfrac{3\sqrt{2}}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{3\sqrt{10}}{5}$

Now you try: $\dfrac{4\sqrt{3}}{\sqrt{7}}=\;?$ Answer: $\dfrac{4\sqrt{21}}{7}$

EXAMPLE 9c — Watch the walkthrough first

Rationalise the denominator: $\dfrac{2\sqrt{7}}{5\sqrt{2}}$.

📺 Step-by-step: multiply by $\dfrac{\sqrt{2}}{\sqrt{2}}$, multiply through, then cancel.

Multiply top and bottom by $\sqrt{2}$: $\dfrac{2\sqrt{7}}{5\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{14}}{5\times 2}=\dfrac{2\sqrt{14}}{10}$
Cancel the common factor of $2$: $\dfrac{2\sqrt{14}}{10}=\boxed{\dfrac{\sqrt{14}}{5}}$

⚠️ Always check at the end: can the numerator and denominator be reduced? Here, $\dfrac{2}{10}=\dfrac{1}{5}$.

Now you try: $\dfrac{2\sqrt{5}}{3\sqrt{2}}=\;?$ Answer: $\dfrac{\sqrt{10}}{3}$ (after cancelling $\dfrac{2}{6}=\dfrac{1}{3}$)

EXAMPLE 9d — Bonus: rationalise a binomial numerator

Rationalise the denominator: $\dfrac{1-\sqrt{3}}{\sqrt{3}}$.

Multiply top and bottom by $\sqrt{3}$:

$\dfrac{1-\sqrt{3}}{\sqrt{3}}=\dfrac{1-\sqrt{3}}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}-3}{3}$

⚠️ Be careful: the $\sqrt{3}$ multiplies both terms on the top (distributive law) — $1\times\sqrt{3}=\sqrt{3}$ and $\sqrt{3}\times\sqrt{3}=3$.

Now you try: $\dfrac{2-\sqrt{7}}{\sqrt{7}}=\;?$ Answer: $\dfrac{2\sqrt{7}-7}{7}$

Practice — Rationalise the denominator and simplify if possible:

$\dfrac{1}{\sqrt{5}}$
$\dfrac{4}{\sqrt{2}}$
$\dfrac{\sqrt{3}}{\sqrt{2}}$
$\dfrac{5}{2\sqrt{3}}$
$\dfrac{3\sqrt{5}}{4\sqrt{2}}$
$\dfrac{2+\sqrt{5}}{\sqrt{5}}$

a) $\dfrac{\sqrt{5}}{5}$
b) $\dfrac{4\sqrt{2}}{2}=2\sqrt{2}$
c) $\dfrac{\sqrt{6}}{2}$
d) $\dfrac{5\sqrt{3}}{2\times 3}=\dfrac{5\sqrt{3}}{6}$
e) $\dfrac{3\sqrt{10}}{4\times 2}=\dfrac{3\sqrt{10}}{8}$
f) $\dfrac{(2+\sqrt{5})\sqrt{5}}{5}=\dfrac{2\sqrt{5}+5}{5}$

Part 3 — Quick quiz (5 min)

Working program — Cambridge Ex 4D

After the quiz, open Cambridge Chapter 4 and complete the following:

Section	Foundation	Standard (set work)	Advanced
Fluency 1–4	1–4 (½)	1–4 (½)	1–4 (⅓)
Problem-solving 5–7	6	5 (½), 6	5–7 (⅓)
Reasoning 8–10	8	8, 9 (½)	9 (⅓), 10
Enrichment 11	—	—	11 (½)

If you finish early, revise Ex 4C using the other lesson page, or start the Chapter 4 review questions.

Exit ticket — write in your book

Before you pack up, in your exercise book write one sentence each:

What number do we multiply $\dfrac{3}{\sqrt{5}}$ by, top and bottom, to rationalise it?
Why doesn't multiplying by $\dfrac{\sqrt{5}}{\sqrt{5}}$ change the value of a fraction?
What should you always check at the end of a rationalising question?