Rationalising a two term denominator | Maths | Quizalize & Oak National Academy

Starter quiz

Expand the following expression: $(3\sqrt {2} + \sqrt {3})(3\sqrt {2} - \sqrt {3})$

'15' ✓
Expand the following expression: $(\sqrt {6} + 1)(\sqrt {6} - 1)$

'5' ✓
Expand the following expression: $(2\sqrt {5} - 3\sqrt {3})(2\sqrt {5} + 3\sqrt {3})$

'-7' ✓
Expand the following expression: $(5\sqrt {7} - 4\sqrt {2})(5\sqrt {7} + 4\sqrt {2})$

'143' ✓
Expand the following expression: $(\sqrt {8} + \sqrt {2})(\sqrt {2} - \sqrt {8})$

'-6' ✓
Expand the following expression: $(\sqrt {10} + 2\sqrt {5})(2\sqrt {5} - \sqrt {10})$

'10' ✓

Match the brackets and their expanded form.

12

⇔

$(\sqrt {8}+2\sqrt {5})(2\sqrt {5}-\sqrt {8})$ ✓

$28+8\sqrt {10}$

⇔

$(\sqrt {8}+2\sqrt {5})(2\sqrt {5}+\sqrt {8})$ ✓

-12

⇔

$(\sqrt {8}-2\sqrt {5})(2\sqrt {5}+\sqrt {8})$ ✓

$-28+8\sqrt {10}$

⇔

$(\sqrt {8}-2\sqrt {5})(2\sqrt {5}-\sqrt {8})$ ✓
You can use the ______ to make an integer from a surd expression.

'difference of two squares' ✓
What would you multiply $(13 + \sqrt {17})$ by to make an integer?

$(13 - \sqrt {17})$ ✓

$(\sqrt {17} + 13)$

$(\sqrt {17} - 13)$ ✓

$(-\sqrt {17} - 13)$
Simplify ${{8} \over {4 - \sqrt {5}}}$

${{32 - 8\sqrt {5}} \over {16 + 5}}$

${{32 - 8\sqrt {5}} \over {21}}$

${{32 + 8\sqrt {5}} \over {16 - 5}}$

${{32 + 8\sqrt {5}} \over {11}}$ ✓
Express ${6} \over {2\sqrt {3} - 4}$ in the form $a(d+\sqrt {3})$ . What is the value of $a$ ?

-6

-3 ✓

2

3
Express ${6} \over {2\sqrt {3} - 4}$ in the form $a(d+\sqrt {3})$ . What is the value of $d$ ?

-6

-3

2 ✓

3

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Fractions should be given in their simplest term.
The denominator should be written as simply as possible.
Combining your knowledge of equivalent fractions and the distributive law will help.

Pupils may struggle to see why using the difference of two squares is important.

The grid method for multiplication can be used to explore why two of the terms cancel each other.

Surd - A surd is an irrational number expressed as the root of a rational number.