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Surds (Cambridge (CIE) IGCSE Maths)

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FrontSimplifying Surds
What is a surd?

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What is a surd?
A surd is a square root of a non-square positive integer.
It is an irrational number.
Examples include $\sqrt{7}, \sqrt{12}, \sqrt{99}$ .
Simplify $\sqrt{a} \times \sqrt{b}$ .
$\sqrt{a} \times \sqrt{b}$ simplifies to $\sqrt{a \times b}$ .
Simplify $\frac{\sqrt{a}}{\sqrt{b}}$ .
$\frac{\sqrt{a}}{\sqrt{b}}$ simplifies to $\sqrt{\frac{a}{b}}$ .
Simplify ${(\sqrt{a})}^{2}$ .
${(\sqrt{a})}^{2}$ simplifies to $a$ .
True or False?
$\sqrt{2} + \sqrt{3}$ simplifies to $\sqrt{5}$ .
False.
$\sqrt{2} + \sqrt{3}$ does not simplify to $\sqrt{5}$ .
The surds need to be of the same type (e.g. all multiple of $\sqrt{2}$ ) to be able to add or subtract.
How can you simplify a surd such as $\sqrt{8}$ ?
To simplify a surd, write the number as a product involving a square number.
Split the surd into the product of two surds, one of which should just become an integer.
E.g. $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2}$ .
How do you add or subtract two surds?
E.g. $\sqrt{8} + \sqrt{18}$ .
To add or subtract two surds, simplify both surds so that they are both multiples of the same surd.
You can then add or subtract by collecting like terms.
For example, $\sqrt{8} + \sqrt{18} = 2 \sqrt{2} + 3 \sqrt{2} = 5 \sqrt{2}$
How do you expand two brackets with expressions containing surds?
E.g. $(2 + \sqrt{3}) (4 + \sqrt{2})$
To expand two brackets with expressions containing surds, you multiply each term in one bracket by each term in the other bracket.
E.g. $\begin{array}{rcl} (2 + \sqrt{3}) (4 + \sqrt{2}) \end{array}$
$\begin{array}{rcl} = & 2 \times 4 + 2 \times \sqrt{2} + \sqrt{3} \times 4 + \sqrt{3} \times \sqrt{2} \\ = & 8 + 2 \sqrt{2} + 4 \sqrt{3} + \sqrt{6} \end{array}$
True or False?
${(a + \sqrt{b})}^{2} = a^{2} + b$ .
False.
${(a + \sqrt{b})}^{2} \neq a^{2} + b$ .
You need to expand using double brackets, $(a + \sqrt{b}) (a + \sqrt{b})$ .
This expands and simplifies to $a^{2} + b + 2 a \sqrt{b}$ .
Simplify $(a + \sqrt{b}) (a - \sqrt{b})$ .
$(a + \sqrt{b}) (a - \sqrt{b})$ simplifies to $a^{2} - b$ .
This is an example of the difference of two squares.
State what is meant by rationalising the denominator.
Rationalising the denominator changes a fraction with surds in the denominator into an equivalent fraction where there are no surds in the denominator.
E.g. The fraction $\frac{2}{\sqrt{5}}$ becomes $\frac{2 \sqrt{5}}{5}$ .
What should you multiply the numerator and the denominator of $\frac{3}{\sqrt{2}}$ by to rationalise the denominator?
To rationalise the denominator of $\frac{3}{\sqrt{2}}$ , you should multiply the numerator and the denominator by $\sqrt{2}$ .
What should you multiply the numerator and the denominator of $\frac{3}{5 - \sqrt{2}}$ by to rationalise the denominator?
To rationalise the denominator of $\frac{3}{5 - \sqrt{2}}$ , you should multiply the numerator and the denominator by $5 + \sqrt{2}$ .
True or False?
Multiplying the numerator and denominator of $\frac{2 + \sqrt{3}}{3 - \sqrt{5}}$ by $2 - \sqrt{3}$ will rationalise the denominator.
False.
Multiplying the numerator and denominator of $\frac{2 + \sqrt{3}}{3 - \sqrt{5}}$ by $2 - \sqrt{3}$ will not rationalise the denominator.
You need to multiply by $3 + \sqrt{5}$ instead.
The multiplier should contain the same terms as the denominator but with the sign between them inverted.
True or False?
Multiplying the numerator and denominator of $\frac{5}{\sqrt{7} - \sqrt{2}}$ by $\sqrt{7} - \sqrt{2}$ will rationalise the denominator.
False.
Multiplying the numerator and denominator of $\frac{5}{\sqrt{7} - \sqrt{2}}$ by $\sqrt{7} - \sqrt{2}$ will not rationalise the denominator.
You should multiply by $\sqrt{7} + \sqrt{2}$ instead.
The sign in between the two terms should be inverted.
To rationalise a denominator $(p + \sqrt{q})$ , you multiply the numerator and the denominator by what expression?
To rationalise a denominator $(p + \sqrt{q})$ , you multiply the numerator and the denominator by $(p - \sqrt{q})$ .

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