Define rationalising the denominator .

Rationalising the denominator means rewriting a fraction as an equivalent one whose denominator is a rational number rather than a surd. The value of the fraction is unchanged; only its form is different, with any surds now sitting in the numerator .

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Surds (Edexcel IGCSE Further Pure Maths): Flashcards

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FrontSimplifying Surds
Define surd.

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Define surd.
A surd is the square root of a non-square integer, such as $\sqrt{2}$ , $\sqrt{13}$ or $\sqrt{99}$ .
Its exact value cannot be written as a whole number or a fraction, which is why $\sqrt{16}$ is not a surd: it is simply $4$ .
Why leave an answer as $5 \sqrt{2}$ rather than $7.071067811 \dots$ ?
$5 \sqrt{2}$ is the exact value, while any decimal version has been rounded and is therefore slightly wrong.
Working in surds all the way through a calculation keeps every step exact, so small rounding errors cannot build up.
Complete the two rules for multiplying and dividing surds:
$\sqrt{a} \times \sqrt{b} = \sqrt{______} and \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{______}$
The completed rules are:
$\sqrt{a} \times \sqrt{b} = \sqrt{a b} and \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
Multiplication and division can be carried out under a single square root, so $\sqrt{3} \times \sqrt{5} = \sqrt{15}$ and $\sqrt{21} \div \sqrt{7} = \sqrt{3}$ .
True or False?
$\sqrt{a} + \sqrt{b}$ can always be simplified to $\sqrt{a + b}$ .
False.
Numbers settle it at once: $\sqrt{9} + \sqrt{4} = 3 + 2 = 5$ , whereas $\sqrt{9 + 4} = \sqrt{13} = 3.6055 \dots$
Subtraction behaves the same way, so $\sqrt{a} - \sqrt{b}$ is not $\sqrt{a - b}$ either.
How do you write $\sqrt{48}$ in its simplest surd form?
Split off the greatest square factor and take its root: $\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4 \sqrt{3}$ .
The number left under the root must have no square factors of its own.
When simplifying a surd, what if you do not spot the greatest square factor first time?
Simply carry on: take out whatever square factor you did find, then check the number left underneath for square factors of its own.
Using $9$ gives $\sqrt{450} = 3 \sqrt{50}$ , and $50$ still has the square factor $25$ , so $3 \sqrt{50} = 15 \sqrt{2}$ .
Why must $\sqrt{32}$ and $\sqrt{8}$ be simplified before they can be added?
Because only like surds can be collected, in just the way that like terms in algebra are collected.
Once simplified both turn out to be multiples of $\sqrt{2}$ : $\sqrt{32} = 4 \sqrt{2}$ and $\sqrt{8} = 2 \sqrt{2}$ , so the sum is $6 \sqrt{2}$ .
Define rationalising the denominator.
Rationalising the denominator means rewriting a fraction as an equivalent one whose denominator is a rational number rather than a surd.
The value of the fraction is unchanged; only its form is different, with any surds now sitting in the numerator.
How do you rationalise the denominator of $\frac{a}{\sqrt{b}}$ ?
Multiply the top and the bottom by $\sqrt{b}$ , the surd that is in the denominator.
The denominator then becomes $\sqrt{b} \times \sqrt{b} = b$ , which is rational, so $\frac{a}{\sqrt{b}} = \frac{a \sqrt{b}}{b}$ .
True or False?
$\frac{1}{\sqrt{5}}$ and $\frac{\sqrt{5}}{5}$ are equal.
True.
Multiplying a fraction by something equal to $1$ leaves its value unchanged, however different the result is made to look.
Both expressions come to $0.4472 \dots$ , so rationalising changes only the form of a number, never the number itself.
What do you multiply by to rationalise $\frac{2}{1 + \sqrt{3}}$ ?
Multiply the top and the bottom by $1 - \sqrt{3}$ , the same expression with the sign in the middle changed.
That expression is called the conjugate of the denominator.
Complete the result that makes the conjugate method work, filling in the missing index and term:
$(a + \sqrt{b}) (a - \sqrt{b}) = a^{______} -______$
The completed result is:
$(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - b$
This is the difference of two squares: the middle terms $- a \sqrt{b}$ and $+ a \sqrt{b}$ cancel each other, and that is what removes the surd.
Write $\frac{4}{\sqrt{6} - 2}$ in the form $p + q \sqrt{r}$ .
Multiply top and bottom by the conjugate $\sqrt{6} + 2$ :
$\frac{4 (\sqrt{6} + 2)}{(\sqrt{6} - 2) (\sqrt{6} + 2)} = \frac{4 (\sqrt{6} + 2)}{6 - 4} = 2 (\sqrt{6} + 2)$
Expanding gives $4 + 2 \sqrt{6}$ , so $p = 4$ , $q = 2$ and $r = 6$ .

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