Surds — Rationalising the Denominator (Ex 4D)
Year 10 Mathematics Core · Cambridge Ch 4, §4D · Mr Wong
Learning intentions.
- Understand that $\sqrt{x}\times\sqrt{x}=x$, so a surd multiplied by itself becomes a whole number
- Know that rationalising the denominator means turning an irrational denominator into a rational one
- Rationalise denominators of the form $\sqrt{y}$ and $b\sqrt{y}$, simplifying fully
Key result
$\dfrac{x}{\sqrt{y}}\;=\;\dfrac{x}{\sqrt{y}}\;\times\;\dfrac{\sqrt{y}}{\sqrt{y}}\;=\;\dfrac{x\sqrt{y}}{y}$
$\dfrac{\sqrt{y}}{\sqrt{y}}=1$, so multiplying by it doesn't change the value — only the form.
Worked example (follow along)
$\dfrac{2\sqrt{7}}{5\sqrt{2}}$ — multiply top & bottom by $\sqrt{2}$:
$=\;\dfrac{2\sqrt{7}}{5\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\;=\;\dfrac{2\sqrt{14}}{5\times 2}\;=\;\dfrac{2\sqrt{14}}{10}\;=\;\dfrac{\sqrt{14}}{5}$ ✓
Always check at the end — can the numerator and denominator be reduced? Here $\dfrac{2}{10}=\dfrac{1}{5}$.
Warm-up
1. Simplify each fraction (no rationalising yet).
a $\dfrac{\sqrt{6}}{\sqrt{6}}=$
b $\dfrac{2\sqrt{5}}{4\sqrt{5}}=$
c $-\dfrac{\sqrt{8}}{\sqrt{2}}=$
d $\dfrac{\sqrt{72}}{\sqrt{2}}=$
2. State the missing number so that the product is a whole number.
a $\sqrt{3}\times\underline{\phantom{xx}}=3$
b $\sqrt{10}\times\sqrt{10}=\underline{\phantom{xx}}$
c $2\sqrt{5}\times\underline{\phantom{xx}}=10$
d $\underline{\phantom{xx}}\times 3\sqrt{7}=21$
Part A — Rationalise (single-term numerator)
3. Rationalise the denominator, simplifying as far as possible.
a $\dfrac{1}{\sqrt{5}}=$
b $\dfrac{4}{\sqrt{2}}=$
c $\dfrac{\sqrt{3}}{\sqrt{2}}=$
d $\dfrac{2\sqrt{7}}{5\sqrt{2}}=$
Surds 4D — continued
Mixed practice · binomial numerators · challenge
Part A — continued
4. Rationalise and simplify.
a $\dfrac{5}{2\sqrt{3}}=$
b $\dfrac{3\sqrt{5}}{4\sqrt{2}}=$
c $\dfrac{6}{\sqrt{12}}=$
d $\dfrac{4\sqrt{3}}{\sqrt{7}}=$
Part B — Binomial numerator
5. Rationalise. Remember to distribute over both terms on the top.
a $\dfrac{1-\sqrt{3}}{\sqrt{3}}$
b $\dfrac{2+\sqrt{5}}{\sqrt{5}}$
c $\dfrac{3+\sqrt{2}}{\sqrt{2}}$
d $\dfrac{2-\sqrt{7}}{\sqrt{7}}$
Challenge
6. Show that $\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{8}}$ simplifies to a single rationalised surd, and find its exact value. Hint: simplify $\sqrt{8}$ first.
7. A right-angled triangle has legs of length $1$ cm and $\sqrt{3}$ cm. (a) Find the length of the hypotenuse. (b) Find the value of $\dfrac{1}{\text{hypotenuse}}$ with a rationalised denominator.