Surds, binomial products and rationalising

Rational and irrational numbers, simplifying and operating with surds, expanding binomial products with surds, conjugates, and rationalising the denominator.

Rational and irrational

A rational number can be written as $\dfrac{a}{b}$ with whole numbers $a$, $b$ (so terminating or recurring decimals). Irrational numbers, such as $\sqrt{2}$ and $\pi$, have non-recurring, never-ending decimals.

Binomial products

Example 1●●

Expand $(\sqrt{5} + 1)^2$ and $(2 + \sqrt{3})(2 - \sqrt{3})$.

$5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5}$

$4 - 3 = 1$ (difference of two squares)

Rationalising the denominator

Example 2●●●

Rationalise $\dfrac{6}{\sqrt{3}}$ and $\dfrac{1}{\sqrt{5} - 2}$.

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

$\dfrac{\sqrt{5} + 2}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \dfrac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2$

Flashcards

Is $0.overline{3}$ rational?

Yes, it equals $\tfrac{1}{3}$.

Conjugate of $sqrt{7} + 2$

$\sqrt{7} - 2$

$(sqrt{a} + sqrt{b})(sqrt{a} - sqrt{b})$

$a - b$

Practice

Expand $(\sqrt{5} + 1)^2$.
Expand $(2 + \sqrt{3})(2 - \sqrt{3})$.
Rationalise $\dfrac{6}{\sqrt{3}}$.
Rationalise $\dfrac{1}{\sqrt{5} - 2}$.
Which is irrational?
Expand $(\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5})$.
Rationalise $\dfrac{\sqrt{2}}{\sqrt{6}}$.
Expand $\sqrt{2}(3 + \sqrt{8})$.