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Some questions on working with surds.

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Manipulate surds and rationalise the denominator of a fraction when it is a surd.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

To include a square root sign in your answer use sqrt(). For example, to write $\\sqrt{3}$, type sqrt(3) into the answer box. If you are entering a number multiplied by the square root of some other number, for example $3\\sqrt{5}$, type 3*sqrt(5) into the answer box.

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a)

Surds can be manipulated using the rule

\\[\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}.\\]

We are asked to state which of $\\sqrt{\\var{p}}$, $\\sqrt{\\simplify{{a}*{n}^2}}$, and $\\sqrt{\\var{a}}$ can be simplified further. Commonly, surds can be simplified if the number inside of the square root has a square number as a factor.

Here, $\\var{p}$ is a prime number which means that its only divisors are $\\var{p}$ and $1$.

Therefore, $\\sqrt{\\var{p}}$ cannot be simplified any further.

Similarly, $\\var{a}$ is also a prime number, so $\\sqrt{\\var{a}}$ also cannot be simplified any further.

On the other hand, $\\simplify{{a}*{n}^2}$ is not a prime number and we can use the previous rule to simplify $\\sqrt{\\simplify{{a}*{n}^2}}$ as

\\[
\\begin{align}
\\sqrt{\\simplify{{a}*{n}^2}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{a}}\\\\
&= \\simplify{{n}*sqrt({a})}.
\\end{align}
\\]

b)

Using the same rule of manipulation as in part a), we can simplify $\\sqrt{\\simplify{{n}^2*{p}}}$ as

\\[
\\begin{align}
\\sqrt{\\simplify{{n}^2*{p}}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{p}}\\\\
&= \\simplify{{n}*sqrt({p})}.
\\end{align}
\\]

c)

Here, we can use both of the rules for manipulating surds:

\\[\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b} \\text{.} \\]

\\[ \\sqrt{\\frac{a}{b}} = \\frac{\\sqrt{a}}{\\sqrt{b}} \\text{.} \\]

We can simplify $\\displaystyle\\frac{ \\sqrt{\\simplify{{a}*{v}}} }{ \\sqrt{\\var{a}} }$ as follows.

\\[
\\begin{align}
\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} &= \\frac{\\sqrt{\\var{a}} \\times \\sqrt{\\var{v}}}{\\sqrt{\\var{a}}} \\\\[0.5em]
&= \\frac{\\sqrt{\\var{a}}}{\\sqrt{\\var{a}}} \\times \\sqrt{\\var{v}} \\\\[0.5em]
&= \\simplify{{sqrt(a)/sqrt(a)}} \\times \\sqrt{\\var{v}} \\\\[0.5em]
&= \\sqrt{\\var{v}} \\text{.}
\\end{align}
\\]

Or,

\\[
\\begin{align}
\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} &= \\sqrt{\\frac{\\simplify{{a}*{v}}}{\\var{a}}} \\\\[0.5em]
&= \\sqrt{\\var{v}} \\text{.}
\\end{align}
\\]

d)

We can simplify the fraction as

\\[
\\begin{align}
\\frac{\\sqrt{\\simplify{({b}{m})^2*{s}}}}{\\var{m}} &= \\frac{\\sqrt{\\simplify{({b*m})^2}} \\times \\sqrt{\\var{s}}}{\\var{m}} \\\\[0.5em]
&= \\frac{\\simplify{{b*m}} \\times \\sqrt{\\var{s}}}{\\var{m}} \\\\[0.5em]
&= \\simplify{{b}*sqrt({s})} \\text{.}
\\end{align}
\\]

e)

\\[
\\begin{align}
\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2{a})+{n}sqrt({b}^2*{a})} &= \\var{d}\\sqrt{\\var{a}} - \\var{b}(\\sqrt{\\simplify{{v}^2}} \\times \\sqrt{\\var{a}})+\\var{n}(\\sqrt{\\simplify{{b}^2}} \\times \\sqrt{\\var{a}}) \\\\
&= \\var{d}\\sqrt{\\var{a}} -\\var{b}(\\simplify{{v}*sqrt({a})})+\\var{n}(\\simplify{{b}*sqrt({a})}) \\\\
&= \\simplify{{d}sqrt({a})}-\\simplify{{b}*{v}sqrt({a})}+\\simplify{{n}*{b}sqrt({a})} \\\\
&= \\simplify{({d}-{b}*{v}+{n}*{b})sqrt({a})} \\text{.}
\\end{align}
\\]

f)

We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{\\sqrt{a}}$, by multiplying the top and bottom by $\\sqrt{a}$.

Therefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$, we multiply top and bottom by $\\sqrt{\\var{a}}$.

\\[
\\begin{align}
\\frac{1}{\\sqrt{\\var{a}}} &= \\frac{1}{\\sqrt{\\var{a}}} \\times \\frac{\\sqrt{\\var{a}}}{\\sqrt{\\var{a}}} \\\\[0.5em]
&= \\frac{\\sqrt{\\var{a}}}{\\var{a}} \\text{.}
\\end{align}
\\]

g)

We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$ by multiplying the top and bottom by $a-\\sqrt{b}$.

Therefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$, we multiply the top and bottom by $\\var{n} - \\sqrt{\\var{a}}$.

\\[
\\begin{align}
\\frac{1}{\\var{n}+\\sqrt{\\var{a}}} &= \\frac{1}{\\var{n}+\\sqrt{\\var{a}}} \\times \\frac{\\var{n}-\\sqrt{\\var{a}}}{\\var{n}-\\sqrt{\\var{a}}} \\\\[0.5em]
&=\\frac{\\var{n}-\\sqrt{\\var{a}}}{(\\var{n}+\\sqrt{\\var{a}})(\\var{n}-\\sqrt{\\var{a}})} \\\\[0.5em]
&=\\frac{\\var{n}-\\sqrt{\\var{a}}}{\\simplify{{n}^2}-\\var{a}} \\\\[0.5em]
&=\\frac{\\var{n}-\\sqrt{\\var{a}}}{\\simplify{{n}^2-{a}}} \\text{.}
\\end{align}
\\]

h)

We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{a-\\sqrt{b}}$ by multiplying the top and bottom by $a+\\sqrt{b}$.

Therefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}}$, we multiply the top and bottom by $\\var{d+p}+\\sqrt{\\var{p}}$.

\\[
\\begin{align}
\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} &= \\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} \\times \\frac{\\var{d+p}+\\sqrt{\\var{p}}}{\\var{d+p}+\\sqrt{\\var{p}}} \\\\[0.5em]
&=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{(\\var{d+p}-\\sqrt{\\var{p}})(\\var{d+p}+\\sqrt{\\var{p}})} \\\\[0.5em]
&=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{\\simplify{{d+p}^2}-\\var{p}} \\\\[0.5em]
&=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{\\simplify{{d+p}^2-{p}}} \\\\[0.5em]
&=\\simplify{{t}/{(d+p)^2-p}}(\\var{d+p}+\\sqrt{\\var{p}}) \\\\[0.5em]
&= \\simplify[all,!noleadingMinus]{({t*(d+p)}+{t}*sqrt({p}))/({(d+p)^2-p})} \\text{.}
\\end{align}
\\]

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Parts c and e

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Fraction in answer for part d.

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parts b and d

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Parts a, d,e and h

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parts d and e

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Which of the following can be simplified further?

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Can be simplified further

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Cannot be simplified further

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$\\sqrt{\\var{p}}$

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$\\sqrt{\\simplify{{a}*{n}^2}}$

", "

$\\sqrt{\\var{a}}$

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Simplify $\\sqrt{\\simplify{{n}^2*{p}}}$.

$\\sqrt{\\simplify{{n}^2*{p}}} =$ [[0]]$\\sqrt{\\var{p}}$.

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Recall the first rule of surds

$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.

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Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} =$ [[0]].

You could use either of the following rules:

$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.

$\\displaystyle\\sqrt{\\frac{a}{b}} = \\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}}$.

You must simplify your answer further.

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You must simplify your answer further.

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Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{({b}{m})^2*{s}}}}{\\var{m}}$.

$\\displaystyle\\frac{\\sqrt{\\simplify{({b}*{m})^2*{s}}}}{\\var{m}} =$ [[0]]$\\sqrt{\\var{s}}$.

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Simplify $\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2*{a})+{n}sqrt({b}^2*{a})}$.

$\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2*{a})+{n}sqrt({b}^2*{a})} =$ [[0]].

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Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{1}{\\sqrt{\\var{a}}} =$ [[0]] [[1]] .

To rationalise the denominator of fractions in the form $\\frac{1}{\\sqrt{a}}$, multiply the top and bottom by $\\sqrt{a}$.

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Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}} =$ [[0]] [[1]] .

To rationalise the denominator of fractions in the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$, multiply the top and bottom by $a-\\sqrt{b}$.

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Rationalise the denominator of the fraction $\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}}$.

$\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} =$ [[0]] [[1]] .

To rationalise the denominator of fractions in the form, $\\displaystyle\\frac{1}{a-\\sqrt{b}}$, multiply the top and bottom by ${a+\\sqrt{b}}$.

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Rationalise the denominator with increasingly difficult examples involving compound denominators.

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Rationalising the denominator means removing any surds on the denominator.

Try the following questions to practise rationalising with expressions in different forms.

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a)

The rule that should be used is $\\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}} = \\sqrt{\\frac{a}{b}}$.

$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{{\\var{prime_nums[0]}}}} = \\sqrt{\\var{square_nums[0]}}= \\simplify{{sqrt(square_nums[0])}}$; and

ii)

$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{{\\var{prime_nums[1]}}}} = \\sqrt{\\var{square_nums[1]}} = \\simplify{{sqrt(square_nums[1])}}$.

b)

To rationalise the denominator, you have to use the rule $\\displaystyle\\sqrt{a}\\times\\sqrt{a}=a$.

You are asked to rationalise the denominator of the expression $\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}$.

As ${\\sqrt{\\var{m_lcm}}}$ is the denominator, we must multiply the whole fraction by ${\\sqrt{\\var{m_lcm}}}$.

\\[ \\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}\\times\\frac{\\sqrt{\\var{m_lcm}}}{\\sqrt{\\var{m_lcm}}} \\]

By using the rule $\\sqrt{a}\\times\\sqrt{b}$ = $\\sqrt{ab}$, we can get the numerator as $\\sqrt{\\var{num}}$.

The denominator is ${\\sqrt{\\var{m_lcm}}}\\times{\\sqrt{\\var{m_lcm}}} = {\\var{m_lcm}}$.

So the whole expression is

\\[ \\frac{\\sqrt{\\var{num}}}{\\var{m_lcm}} \\]

Finally, we can cancel down by finding the greatest common divisor of the top and bottom to obtain

\\[ \\frac{\\sqrt{\\var{a[0]*b[0]}}}{\\var{b[0]}} \\]

c)

The first thing to do in this question is multiply the whole fraction by the surd on the denominator, $\\sqrt{\\var{f}}$. Using the rule $\\displaystyle\\sqrt{a}\\times\\sqrt{a}=a$, the new value of the denominator is $\\var{d}\\times\\var{f}=\\var{df}$.

$\\displaystyle\\frac{\\var{c}}{\\var{d}\\sqrt{\\var{f}}}\\times\\simplify[all,!simplifyFractions,!sqrtDivision]{sqrt({f})/(sqrt({f}))}={\\frac{\\simplify{{c}*sqrt({f})}}{{(\\var{d}\\times\\var{f})}}}=\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}$.

Make sure to simplify your fractions down to their simplest form. The fraction can be cancelled down using the highest common divisor, $\\var{gcd_cdf}$, to give a final answer of

\\[\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}=\\frac{\\simplify{{c_coprime}*sqrt({f})}}{\\var{df_coprime}}\\]

We are asked to find an answer in the form $\\frac{a \\sqrt{\\var{f}}}{b}$, so $a = \\var{c_coprime}$ and $b = \\var{df_coprime}$.

d)

When rationalising an expression with a compound denominator, like $\\displaystyle\\frac{\\var{g}}{\\sqrt{\\var{h}}+\\var{j}}$, we must multiply the fraction by the conjugate of the denominator.

The conjugate of $(\\sqrt{a}+b)$ is $(\\sqrt{a}-b)$, and therefore the conjugate of $(\\sqrt{\\var{h}}+\\var{j})$ is $(\\sqrt{\\var{h}}-\\var{j})$.

\\[ \\frac{\\var{g}}{\\sqrt{\\var{h}}+\\var{j}}\\times\\frac{(\\sqrt{\\var{h}}-\\var{j})}{(\\sqrt{\\var{h}}-\\var{j})}=\\frac{\\var{g}(\\sqrt{\\var{h}}-\\var{j})}{(\\sqrt{\\var{h}}+\\var{j})(\\sqrt{\\var{h}}-\\var{j})}\\text{.} \\]

From this point, we need to multiply out the brackets on the numerator and denominator, using the rule $\\displaystyle\\sqrt{a}\\times\\sqrt{a}=a$.

$\\displaystyle\\frac{\\var{g}(\\sqrt{\\var{h}}-\\var{j})}{(\\sqrt{\\var{h}}+\\var{j})(\\sqrt{\\var{h}}-\\var{j})}==\\frac{\\simplify[all,!collectNumbers,!noLeadingMinus]{({-{g}*{j}})+{g}}\\sqrt{\\var{h}}}{\\var{h}{-\\simplify{{j}*sqrt({h})}}+\\simplify{{j}*sqrt({h})}-{{\\var{j^2}}}}$.

The two $\\simplify{{j}*sqrt({h})}$ terms will cancel on the denominator, so that you are left with

\\[ \\frac{ \\simplify{ {g*-j} + {g}*sqrt({h})} }{\\var{h-j^2}}\\text{,} \\]

which simplifies to

\\[ \\var{o}\\sqrt{\\var{h}}-{\\var{l}}\\text{.} \\]

e)

$\\displaystyle\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}}$ =

To rationalise the denominator of this fraction, multiply the whole fraction by the conjugate of the denominator, multiply out the brackets and collect like terms.

$\\displaystyle\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}}\\times\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}-\\sqrt{\\var{u}}}=\\frac{(\\var{t}-\\sqrt{\\var{u}})(\\var{t}-\\sqrt{\\var{u}})}{(\\var{t}+\\sqrt{\\var{u}})(\\var{t}-\\sqrt{\\var{u}})}=\\frac{\\var{t^2}-\\var{2t}\\sqrt{\\var{u}}+\\var{u}}{\\var{t^2}-\\var{u}}$.

$\\displaystyle\\frac{\\var{t^2}-\\var{2t}\\sqrt{\\var{u}}+\\var{u}}{\\var{t^2}-\\var{u}}=\\frac{\\var{t^2+u}-\\var{2t}{\\sqrt{\\var{u}}}}{\\var{t^2-u}}$.

Dividing all terms by their highest common divisor, ${\\var{gcd_frac}}$, gives

$\\displaystyle\\frac{\\var{num_simp_1}-\\simplify{{num_simp_2}*sqrt({u})}}{\\var{denom_simp}}$.

f)

To add $\\displaystyle\\frac{\\var{p}}{\\sqrt{\\var{p}}+\\sqrt{\\var{p^3}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}}$ you must put both terms over the same denominator.

It is good to notice that $\\sqrt{\\var{p^3}}$ can be rewritten as $\\var{p}\\sqrt{\\var{p}}$, meaning we can rewrite the first fraction as

\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\var{p}\\sqrt{\\var{p}}} \\text{.} \\]

This can be simplified further by collecting like terms to obtain

\\[ \\frac{\\var{p}}{\\var{p+1}\\sqrt{\\var{p}}} \\text{.} \\]

Therefore, the expression is now $\\displaystyle\\frac{\\var{p}}{{\\var{p+1}}\\sqrt{\\var{p}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}}$.

In order to complete the addition, the second fraction also has to have the same denominator as the first fraction. We have to multiply it by $\\simplify{{p_p1}sqrt({p})}$ in order to get common denominators across both fractions.

$\\displaystyle\\frac{\\var{r_coprime}}{\\var{s_coprime}}\\times\\simplify[all,!simplifyFractions,!sqrtDivision]{{p_p1}*sqrt({p})/({p_p1}*sqrt({p}))}=\\frac{\\simplify{{one}*sqrt({p})}}{\\var{two}\\sqrt{\\var{p}}}$,

Both parts of the expression now have the same denominator and we can add them.

\\[ \\frac{\\var{p}}{\\var{p+1}\\sqrt{\\var{p}}}+\\frac{\\simplify{{one}*sqrt({p})}}{\\var{two}\\sqrt{\\var{p}}}= \\frac{\\var{p}+\\simplify{{one}*sqrt({p})}}{\\var{p+1}\\sqrt{\\var{p}}} \\]

When simplifying, if we note that $\\displaystyle\\sqrt{a}\\times\\sqrt{a}=a$, we can pull $\\sqrt{\\var{p}}$ out of the numerator as a common factor:

\\[ \\frac{\\sqrt{\\var{p}}(\\sqrt{\\var{p}}+\\var{r_coprime*p_p1})}{\\var{p+1}\\sqrt{\\var{p}}}\\text{.} \\]

We can then cancel the common $\\sqrt{\\var{p}}$ term on the numerator and denominator to simplify the expression further and get a final answer of

\\[ \\frac{\\sqrt{\\var{p}}+\\var{one}}{\\var{p+1}} \\text{.} \\]

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Random number.

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Simplified numerical term of numerator.

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GCD of terms in numerator.

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Simplified denominator.

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Multiple of LCM of n amd m.

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GCD of all terms in fraction.

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Simplified surd term of numerator.

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Rationalise the denominator of the following surds to simplify them down to their integer value.

$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}}=$ [[0]]

ii)

$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}}=$ [[1]]

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Rationalise the denominator and rewrite as a fraction in its simplest form.

\n\n\n\n\n\n\n\n\n\n\n

$\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}=$	$\\surd$[[0]]
[[1]]

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Rationalise the denominator of this expression and reduce to lowest terms.

\\[ \\frac{\\var{c}}{\\var{d}\\sqrt{\\var{f}}} = \\frac{a\\sqrt{\\var{f}}}{b}\\text{,} \\]

where

$a =$ [[0]]

$b =$ [[1]]

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Express the following in the form $a\\sqrt{b}+m$, where $a$, $b$ and $c$ are integers.

$\\displaystyle\\frac{\\var{g}}{(\\sqrt{\\var{h}}+\\var{j})}$ =

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$\\displaystyle\\var{o}\\sqrt{\\var{h}}+\\var{l}$

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$\\displaystyle-\\var{l}-\\var{o}\\sqrt{\\var{h}}$

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$\\displaystyle\\var{o}\\sqrt{\\var{h}}-\\var{l}$

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$\\displaystyle\\var{l}-\\sqrt{\\var{h}}$

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Rationalise the denominator of this expression and reduce down to its simplest form.

\\[ \\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}} = \\frac{a-b\\sqrt{\\var{u}}}{c} \\text{,} \\]

where:

$a = $ [[0]]

$b = $ [[1]]

$c = $ [[2]]

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Complete this addition by firstly rationalising the denominator of the first fraction. Your final answer should be a fraction with integer denominator.

\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\sqrt{\\var{p^3}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}} \\]

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$\\displaystyle\\frac{\\sqrt{\\var{p}}+\\var{r_coprime*p_p1}}{\\var{p+1}}$

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$\\displaystyle\\frac{\\sqrt{\\var{p}}(\\sqrt{\\var{p}}+\\var{r_coprime*p_p1})}{\\var{p+1}\\sqrt{\\var{p}}}$

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$\\displaystyle\\frac{\\var{p}+\\var{qwerty}\\sqrt{\\var{p}}}{\\var{p}\\sqrt{\\var{p}}}$

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$\\displaystyle\\frac{\\sqrt{\\var{p}}+\\var{r_coprime*p_p1}}{\\var{p}}$

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Surd

", "

Not a surd

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$\\sqrt{\\var{square}}$

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$\\sqrt{\\var{h}}$

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$^3\\sqrt{\\var{cube}}$

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$\\sqrt{\\var{j}}$

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$\\sqrt{\\var{k}}$

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For the following examples, tick the correct box to determine whether or not they are a surd.

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$4\\sqrt3$

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$2\\sqrt{11}$

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$2\\sqrt{14}$

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$4\\sqrt{2}$

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i) $\\sqrt{48}$

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ii) $\\sqrt{32}$

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iii) $\\sqrt{56}$

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iv) $\\sqrt{44}$

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Match each surd with the equivalent simplification.

[[0]]

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Simplify the following surds:

$\\displaystyle\\sqrt{\\var{c}}$ = [[0]]$\\displaystyle\\sqrt{\\var{b}}$

$\\displaystyle\\sqrt{\\var{g}}$ = [[1]]$\\displaystyle\\sqrt{\\var{f}}$

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a)

$\\sqrt{\\var{square}}$ and $\\sqrt[3]{\\var{cube}}$ are not surds, as they can be simplified to whole integers: $\\simplify{{sqrt(square)}}$ and $\\var{root}$ respectively. They are roots, but not surds. All surds are roots but not all roots are surds.

$\\sqrt{\\var{h}}$, $\\sqrt{\\var{j}}$ and $\\sqrt{\\var{k}}$ are surds, as they cannot be simplified to a whole integer. There is no number, $b$, such that $b^2=\\var{h}, \\var{j}$ or $\\var{k}$. Therefore, $\\sqrt{\\var{h}}$, $\\sqrt{\\var{j}}$ and $\\sqrt{\\var{k}}$ are both roots and surds.

b)

The rule that should be used is $\\sqrt{a}\\times\\sqrt{b}=\\sqrt{ab}$.

We need to try to find a square number that divides $ab$ and rewrite this as $\\sqrt{b^2}\\times\\sqrt{a}$.

$\\sqrt{48}$ = $\\sqrt{16}\\times\\sqrt3$

$\\sqrt{16}$ simplifies down to $4$ so the final answer is: $4\\sqrt3$.

ii)

$\\sqrt{56}$ = $\\sqrt{4}\\times\\sqrt{14}$

$\\sqrt4$ simplifies down to $2$ so the final answer is: $2\\sqrt{14}$.

iii)

$\\sqrt{32}$ = $\\sqrt{16}\\times\\sqrt{2}$

$\\sqrt{16}$ simplifies down to $4$ so the final answer is: $4\\sqrt2$.

iv)

$\\sqrt{44}$ = $\\sqrt{4}\\times\\sqrt{11}$

$\\sqrt4$ simplifies down to $2$ so the final answer is: $2\\sqrt{11}$.

c)

This question requires you to notice that $\\sqrt{\\var{a}}$ and $\\sqrt{\\var{d}}$ are squared numbers and can be simplified to integers.

$\\sqrt{\\var{a}}$ = $\\var{sqrta}$ such that:

i) $\\sqrt{\\var{c}}$ = $\\sqrt{\\var{a}}$ x $\\sqrt{\\var{b}}$ = $\\var{sqrta}\\sqrt{\\var{b}}$ and

ii) $\\sqrt{\\var{g}}$ = $\\sqrt{\\var{d}}$ x $\\sqrt{\\var{f}}$ = $\\var{sqrtd}\\sqrt{\\var{f}}$.

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a times b

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List of squared numbers from 1 to 144

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square root of the selected square number d.

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square root of the squared numbers

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Random number between 1 and 12 except 4 and 9.

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d times f

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Random squared number but not the same number as a.

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Random number between 2 and 12

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Random squared number

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List of random square number between 1 and 36

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Surds are square roots that cannot be simplified to a whole number. They have a decimal equivalent but their decimal representations are never-ending. Therefore, it is often easier to leave surds as they are in algebraic calculations.

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This question tests the student's understanding of what is and is not a surd, and on their simplification of surds.

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