When simplifying a surd, what kind of factor do you look for under the root, and which one do you choose?

Look for a factor that is a square number , and choose the largest one available. A smaller square factor still gives a correct step, but leaves behind a surd that can be simplified further.

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Surds (SQA National 5 Maths): Flashcards

Exam code: X847 75

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FrontSimplification with Surds
Define surd.

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Cards in this collection (14)

Define surd.
A surd is the square root of an integer that is not a square number, such as $\sqrt{2}$ or $\sqrt{13}$ .
Its value is irrational, so leaving an answer as a surd keeps it exact rather than rounding it to a decimal.
Why does $\sqrt{2} \times \sqrt{8}$ simplify to a whole number, when neither factor is one?
Numbers under square roots can be multiplied together, so $\sqrt{a} \times \sqrt{b} = \sqrt{a b}$ .
That gives $\sqrt{2} \times \sqrt{8} = \sqrt{16}$ , and 16 is a square number, so the answer is the whole number 4.
Complete the rule for dividing one surd by another.
$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{______}$
The completed rule is:
$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
For example $\frac{\sqrt{21}}{\sqrt{7}} = \sqrt{21 \div 7} = \sqrt{3}$ .
True or False?
$\sqrt{9} + \sqrt{4} = \sqrt{13}$
False.
$\sqrt{9} + \sqrt{4} = 3 + 2 = 5$ , whereas $\sqrt{13} = 3.605 \dots$
Numbers under square roots can be multiplied or divided, but never added or subtracted.
When can two surd terms be added or subtracted, and when can they not?
Only like surds combine, meaning terms with the same number under the root, so $3 \sqrt{5} + 8 \sqrt{5} = 11 \sqrt{5}$ .
Terms such as $2 \sqrt{3} + 4 \sqrt{6}$ have different roots and cannot be combined, in the same way that $2 x + 4 y$ cannot be simplified.
When simplifying a surd, what kind of factor do you look for under the root, and which one do you choose?
Look for a factor that is a square number, and choose the largest one available.
A smaller square factor still gives a correct step, but leaves behind a surd that can be simplified further.
Complete the working that writes $\sqrt{720}$ in its simplest surd form.
$\sqrt{720} = \sqrt{______\times 5} =______\sqrt{5}$
The completed working is:
$\sqrt{720} = \sqrt{144 \times 5} = 12 \sqrt{5}$
Splitting the root gives $\sqrt{144} \times \sqrt{5}$ , and $\sqrt{144}$ has the exact value 12.
Which result about surds lets you simplify $\sqrt{7} \times \sqrt{7}$ when expanding brackets?
The result ${(\sqrt{a})}^{2} = a$ means a surd multiplied by itself gives the number under the root.
Expanding $\sqrt{7} (\sqrt{7} - \sqrt{3})$ therefore gives $7 - \sqrt{21}$ .
True or False?
$\sqrt{32} + \sqrt{8}$ can be written as a single surd term.
True.
Simplifying each surd separately gives $\sqrt{32} = 4 \sqrt{2}$ and $\sqrt{8} = 2 \sqrt{2}$ , which are like surds.
They combine to give $6 \sqrt{2}$ .
Define rationalising the denominator.
Rationalising the denominator means rewriting a fraction as an equivalent fraction whose denominator contains no surd.
For example, $\frac{4}{\sqrt{5}}$ is rewritten as $\frac{4 \sqrt{5}}{5}$ .
When a fraction has a single surd as its denominator, what do you multiply it by, and why does that work?
Multiply the top and bottom by that same surd, which is the same as multiplying by 1, so the value of the fraction does not change.
The denominator then becomes $\sqrt{b} \times \sqrt{b} = b$ , which is a rational number.
Complete the general result for rationalising a fraction with a single surd in the denominator.
$\frac{a}{\sqrt{b}} = \frac{a \sqrt{______}}{______}$
The completed result is:
$\frac{a}{\sqrt{b}} = \frac{a \sqrt{b}}{b}$
The surd moves from the denominator up into the numerator.
Express $\frac{8}{\sqrt{14}}$ with a rational denominator, in its simplest form.
Multiplying the top and bottom by $\sqrt{14}$ gives $\frac{8 \sqrt{14}}{14}$ .
Dividing the 8 and the 14 by their common factor of 2 gives the simplest form:
$\frac{4 \sqrt{14}}{7}$
True or False?
The number $\sqrt{\frac{2}{3}}$ needs its denominator rationalising.
True.
The root of a fraction can be split, so $\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$ , which has a surd underneath.
Multiplying the top and bottom by $\sqrt{3}$ gives $\frac{\sqrt{6}}{3}$ .

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