Year 10 Core — Multiplying & Dividing Surds (4C)

Today's lesson

We're continuing Chapter 4 — Surds. By the end of today you should be able to:

Learning intentions

Multiply two surds using $\sqrt{x}\times\sqrt{y}=\sqrt{xy}$
Divide two surds using $\dfrac{\sqrt{x}}{\sqrt{y}}=\sqrt{\dfrac{x}{y}}$
Use the fact that $\sqrt{x}\times\sqrt{x}=x$ to simplify squares of surds
Apply the distributive law to brackets involving surds

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

When you add surds you can only combine like surds (e.g. $3\sqrt{2}+5\sqrt{2}=8\sqrt{2}$). Multiplying is different — it works for any two surds. Watch:

📺 Key rule: why $\sqrt{x}\times\sqrt{y}=\sqrt{xy}$ — with a calculator check.

The two rules

$\sqrt{x}\times\sqrt{y}=\sqrt{xy}$ and $\dfrac{\sqrt{x}}{\sqrt{y}}=\sqrt{\dfrac{x}{y}}$

More generally: $a\sqrt{x}\times b\sqrt{y}=ab\sqrt{xy}$ and $\dfrac{a\sqrt{x}}{b\sqrt{y}}=\dfrac{a}{b}\sqrt{\dfrac{x}{y}}$

And, crucially, $\sqrt{x}\times\sqrt{x}=x$ (a surd squared just removes the root).

Building understanding — state the missing parts:

$\sqrt{15}\div\sqrt{3}=\sqrt{\underline{\phantom{15/3}}}=\sqrt{\underline{\phantom{5}}}$
$\sqrt{42}\div\sqrt{7}=\sqrt{\underline{\phantom{42/7}}}=\sqrt{\underline{\phantom{6}}}$
$\sqrt{6}\times\sqrt{5}=\sqrt{6\times\underline{\phantom{5}}}=\sqrt{\underline{\phantom{30}}}$
$\sqrt{6}\times\sqrt{6}=\underline{\phantom{6}}$ $\sqrt{7}^{\,2}=\underline{\phantom{7}}$ $(\sqrt{5})^{2}=\underline{\phantom{5}}$

a) $\sqrt{15/3}=\sqrt{5}$
b) $\sqrt{42/7}=\sqrt{6}$
c) $\sqrt{6\times 5}=\sqrt{30}$
d) $\sqrt{6}\times\sqrt{6}=6$, $\sqrt{7}^{\,2}=7$, $(\sqrt{5})^{2}=5$

Part 2 — Multiplying surds (Example 6, ~12 min)

EXAMPLE 6a

Simplify $\sqrt{2}\times\sqrt{3}$.

Use the rule $\sqrt{x}\times\sqrt{y}=\sqrt{xy}$:

$\sqrt{2}\times\sqrt{3}=\sqrt{2\times 3}=\sqrt{6}$

Now you try: $\sqrt{5}\times\sqrt{3}=\;?$ Answer: $\sqrt{15}$

EXAMPLE 6b — Watch the walkthrough first

Simplify $2\sqrt{3}\times 3\sqrt{15}$.

📺 Step-by-step: multiply the whole numbers, multiply the radicands, then simplify.

Group whole numbers and surds: $2\sqrt{3}\times 3\sqrt{15}=(2\times 3)\sqrt{3\times 15}=6\sqrt{45}$
Simplify $\sqrt{45}$: $45=9\times 5$, so $\sqrt{45}=3\sqrt{5}$
$=6\times 3\sqrt{5}=\boxed{18\sqrt{5}}$

Now you try: $3\sqrt{2}\times 4\sqrt{6}=\;?$ Answer: $24\sqrt{3}$

EXAMPLE 6c

Simplify $(2\sqrt{5})^{2}$.

$(2\sqrt{5})^{2}=2\sqrt{5}\times 2\sqrt{5}=4\times(\sqrt{5})^{2}=4\times 5=\boxed{20}$

Now you try: $(3\sqrt{7})^{2}=\;?$ Answer: $9\times 7=63$

Practice 2.1 — Multiply and simplify:

$\sqrt{3}\times\sqrt{7}$
$\sqrt{10}\times\sqrt{2}$
$4\sqrt{2}\times\sqrt{5}$
$3\sqrt{2}\times 5\sqrt{3}$
$2\sqrt{6}\times 5\sqrt{2}$
$3\sqrt{5}\times 2\sqrt{10}$

a) $\sqrt{21}$
b) $\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}$
c) $4\sqrt{10}$
d) $15\sqrt{6}$
e) $10\sqrt{12}=10\sqrt{4\times 3}=20\sqrt{3}$
f) $6\sqrt{50}=6\sqrt{25\times 2}=30\sqrt{2}$

Practice 2.2 — Squares of surds:

$(\sqrt{6})^{2}$
$(\sqrt{11})^{2}$
$(3\sqrt{2})^{2}$
$(4\sqrt{5})^{2}$
$(5\sqrt{3})^{2}$

a) $6$
b) $11$
c) $9\times 2=18$
d) $16\times 5=80$
e) $25\times 3=75$

Part 3 — Dividing surds (Example 7, ~10 min)

EXAMPLE 7a

Simplify $-\sqrt{10}\div\sqrt{2}$.

Use $\dfrac{\sqrt{x}}{\sqrt{y}}=\sqrt{\dfrac{x}{y}}$:

$-\sqrt{10}\div\sqrt{2}=-\sqrt{\dfrac{10}{2}}=-\sqrt{5}$

Now you try: $-\sqrt{15}\div\sqrt{5}=\;?$ Answer: $-\sqrt{3}$

EXAMPLE 7b

Simplify $\dfrac{12\sqrt{18}}{3\sqrt{3}}$.

Split the whole numbers from the surds: $\dfrac{12\sqrt{18}}{3\sqrt{3}}=\dfrac{12}{3}\sqrt{\dfrac{18}{3}}$
$=4\sqrt{6}$

Now you try: $\dfrac{14\sqrt{22}}{7\sqrt{11}}=\;?$ Answer: $2\sqrt{2}$

Practice 3.1 — Divide and simplify:

$\sqrt{20}\div\sqrt{5}$
$\sqrt{30}\div\sqrt{6}$
$\sqrt{72}\div\sqrt{8}$
$\dfrac{10\sqrt{24}}{5\sqrt{6}}$
$\dfrac{8\sqrt{45}}{2\sqrt{5}}$
$\dfrac{6\sqrt{50}}{3\sqrt{2}}$

a) $\sqrt{4}=2$
b) $\sqrt{5}$
c) $\sqrt{9}=3$
d) $\dfrac{10}{5}\sqrt{\dfrac{24}{6}}=2\sqrt{4}=4$
e) $4\sqrt{9}=12$
f) $2\sqrt{25}=10$

Part 4 — The distributive law (Example 8, ~12 min)

Same as with letters: $a(b+c)=ab+ac$. Multiply the term outside the bracket through to each term inside, then simplify any surds you can.

EXAMPLE 8a

Expand and simplify $\sqrt{3}(3\sqrt{5}-\sqrt{6})$.

Distribute the $\sqrt{3}$: $\sqrt{3}(3\sqrt{5}-\sqrt{6})=3\sqrt{15}-\sqrt{18}$
Simplify $\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2}$
$=3\sqrt{15}-3\sqrt{2}$

Now you try: $\sqrt{2}(5\sqrt{3}-\sqrt{7})=\;?$ Answer: $5\sqrt{6}-\sqrt{14}$

EXAMPLE 8b

Expand and simplify $3\sqrt{6}(2\sqrt{10}-4\sqrt{6})$.

Distribute: $3\sqrt{6}\times 2\sqrt{10}=6\sqrt{60}$ and $3\sqrt{6}\times 4\sqrt{6}=12\times 6=72$
So $3\sqrt{6}(2\sqrt{10}-4\sqrt{6})=6\sqrt{60}-72$
Simplify $\sqrt{60}=\sqrt{4\times 15}=2\sqrt{15}$
$=6\times 2\sqrt{15}-72=12\sqrt{15}-72$

Now you try: $5\sqrt{3}(2\sqrt{6}-3\sqrt{3})=\;?$ Answer: $30\sqrt{2}-45$

Practice 4.1 — Expand and simplify:

$\sqrt{2}(\sqrt{5}+\sqrt{7})$
$\sqrt{3}(2\sqrt{3}-\sqrt{5})$
$2\sqrt{5}(\sqrt{10}+3\sqrt{5})$
$\sqrt{6}(\sqrt{2}+\sqrt{3})$
$4\sqrt{2}(3\sqrt{8}-\sqrt{5})$

a) $\sqrt{10}+\sqrt{14}$
b) $2\times 3-\sqrt{15}=6-\sqrt{15}$
c) $2\sqrt{50}+6\times 5=2\times 5\sqrt{2}+30=10\sqrt{2}+30$
d) $\sqrt{12}+\sqrt{18}=2\sqrt{3}+3\sqrt{2}$
e) $12\sqrt{16}-4\sqrt{10}=12\times 4-4\sqrt{10}=48-4\sqrt{10}$

Part 5 — Quick quiz (5 min)

Working program — Cambridge Ex 4C

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

Section	Foundation	Standard (set work)	Advanced
Fluency 1–7	1–7 (½)	1–7 (½)	1–7 (⅓)
Problem-solving 8–10	8	8, 9 (½)	8, 9 (½), 10
Reasoning 11–13	11	11, 12	12, 13
Enrichment 14–15	—	—	14–15 (½)

If you finish early, start Ex 4D — Rationalising the Denominator using the other lesson page.

Exit ticket — write in your book

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

What is the rule for multiplying two surds?
Why does $\sqrt{6}\times\sqrt{6}=6$?
What's the first step when you expand $3\sqrt{2}(\sqrt{5}-\sqrt{2})$?