What does it mean to rationalise the denominator of a fraction?

Rationalising the denominator means rewriting the fraction so that the denominator is a whole number , with any surds moved into the numerator. A fraction with a surd on the bottom is not counted as being in its simplest form.

When rationalising, why must you multiply the top and bottom by the same thing, rather than just the bottom?

Multiplying the top and bottom by the same thing is multiplying by 1, so the value of the fraction does not change, only its appearance. Multiplying only the denominator would give a different number altogether, not an equivalent fraction.

GCSEFurther MathsAQAFlashcards1. NumberSurds

Surds (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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FrontSimplifying Surds
Define surd.

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Cards in this collection (15)

Define surd.
A surd is a square root whose exact value cannot be written as a whole number or a fraction.
Leaving an answer as a surd keeps it exact: $5 \sqrt{2}$ rather than $7.071067812$ .
Why is $\sqrt{16}$ not a surd, when $\sqrt{13}$ is?
16 is a square number, so $\sqrt{16} = 4$ exactly.
13 is not a square number, so $\sqrt{13}$ can only be written exactly by keeping the root sign.
True or False?
$\sqrt{9} + \sqrt{4} = \sqrt{13}$
False.
$\sqrt{9} + \sqrt{4} = 3 + 2 = 5$ , but $\sqrt{13} = 3.605 \dots$ , so the two are not equal.
You cannot add the numbers underneath the roots; you can only collect like surds, in the same way you collect $3 x$ and $8 x$ : $3 \sqrt{5} + 8 \sqrt{5} = 11 \sqrt{5}$ .
Complete the rules for multiplying and dividing surds:
$\sqrt{a} \times \sqrt{b} = \sqrt{______}$
$\sqrt{a} \div \sqrt{b} = \sqrt{______}$
The completed rules are:
$\sqrt{a} \times \sqrt{b} = \sqrt{a b}$
$\sqrt{a} \div \sqrt{b} = \sqrt{\frac{a}{b}}$
Both work in reverse, which is what lets you factorise a surd: $\sqrt{35} = \sqrt{5} \times \sqrt{7}$ .
True or False?
Multiplying two surds together can give a whole number.
True.
$\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$ , so two surds have multiplied to give an integer.
An expression containing surds will not always have a surd in its final answer.
How do you write $\sqrt{72}$ in its simplest surd form?
Split off the largest square factor and take its root:
$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6 \sqrt{2}$
A smaller square factor still works but leaves more to do: $\sqrt{4 \times 18} = 2 \sqrt{18}$ is not yet in simplest form.
How do you simplify $\sqrt{32} + \sqrt{8}$ ?
Simplify each surd separately first, so that the same surd appears in both terms:
$\sqrt{32} + \sqrt{8} = 4 \sqrt{2} + 2 \sqrt{2} = 6 \sqrt{2}$
Simplifying is what reveals the like terms; before it, $\sqrt{32}$ and $\sqrt{8}$ look unrelated.
Expand and simplify ${(3 + \sqrt{5})}^{2}$ .
Expand as you would any double bracket, then use ${(\sqrt{a})}^{2} = a$ :
${(3 + \sqrt{5})}^{2} = 9 + 3 \sqrt{5} + 3 \sqrt{5} + 5 = 14 + 6 \sqrt{5}$
Squaring the surd is what produces the extra whole number, here the 5 that turns 9 into 14.
Write $\sqrt{200} - \sqrt{72} + 3 \sqrt{162}$ in the form $a \sqrt{2}$ .
The answer is $31 \sqrt{2}$ .
The only awkward term is $3 \sqrt{162}$ : since $\sqrt{162} = 9 \sqrt{2}$ , that 9 multiplies the 3 already outside the root to give $27 \sqrt{2}$ .
With $\sqrt{200} = 10 \sqrt{2}$ and $\sqrt{72} = 6 \sqrt{2}$ , the coefficients give $10 - 6 + 27 = 31$ .
What does it mean to rationalise the denominator of a fraction?
Rationalising the denominator means rewriting the fraction so that the denominator is a whole number, with any surds moved into the numerator.
A fraction with a surd on the bottom is not counted as being in its simplest form.
Complete the identity that makes rationalising work:
$(a + \sqrt{b}) (a - \sqrt{b}) =______-______$
The completed identity is:
$(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - b$
The two middle terms, $- a \sqrt{b}$ and $+ a \sqrt{b}$ , cancel, which is why no surd survives.
When rationalising, why must you multiply the top and bottom by the same thing, rather than just the bottom?
Multiplying the top and bottom by the same thing is multiplying by 1, so the value of the fraction does not change, only its appearance.
Multiplying only the denominator would give a different number altogether, not an equivalent fraction.
How do you decide what to multiply the top and bottom by when rationalising?
If the denominator is a single surd, multiply by that surd: for $\frac{5}{\sqrt{3}}$ , use $\sqrt{3}$ .
If the denominator has two terms, multiply by its conjugate, the same two terms with the sign between them changed: for $2 + \sqrt{5}$ , use $2 - \sqrt{5}$ .
True or False?
After rationalising, the denominator is always a positive whole number.
False.
The denominator only has to be a whole number, and it can be negative: $(2 - \sqrt{7}) (2 + \sqrt{7}) = 4 - 7 = - 3$ .
Divide through by the negative as normal; the minus sign simply moves into the numerator.
Rationalise $\frac{3 \sqrt{2} + 4}{5 \sqrt{2} - 7}$ .
Multiply top and bottom by $5 \sqrt{2} + 7$ :
$\frac{(3 \sqrt{2} + 4) (5 \sqrt{2} + 7)}{(5 \sqrt{2} - 7) (5 \sqrt{2} + 7)} = \frac{30 + 21 \sqrt{2} + 20 \sqrt{2} + 28}{50 - 49} = 58 + 41 \sqrt{2}$
The denominator comes to 1, so the whole fraction disappears and the numerator is the answer.

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