// Numbas version: exam_results_page_options {"name": "Simon's copy of Rationalising the denominator - surds", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variables": {"c_coprime": {"description": "", "templateType": "anything", "definition": "c/gcd_cdf", "name": "c_coprime", "group": "Part c"}, "h": {"description": "", "templateType": "anything", "definition": "j^2+random(-3..3 except 0)", "name": "h", "group": "Part d"}, "df_coprime": {"description": "", "templateType": "anything", "definition": "df/gcd_cdf", "name": "df_coprime", "group": "Part c"}, "g": {"description": "", "templateType": "anything", "definition": "k*(h-j^2)", "name": "g", "group": "Part d"}, "df": {"description": "", "templateType": "anything", "definition": "d*f", "name": "df", "group": "Part c"}, "t": {"description": "", "templateType": "anything", "definition": "random(3..6)", "name": "t", "group": "Part e"}, "num": {"description": "", "templateType": "anything", "definition": "m_lcm*n_lcm", "name": "num", "group": "Part b"}, "p_p1": {"description": "", "templateType": "anything", "definition": "ceil(p/(s_coprime))", "name": "p_p1", "group": "Part f"}, "d": {"description": "", "templateType": "anything", "definition": "random(2..10)", "name": "d", "group": "Part c"}, "u": {"description": "", "templateType": "anything", "definition": "random(2..8 except 4 except t^2-1) ", "name": "u", "group": "Part e"}, "prime_nums": {"description": "

List of prime numbers.

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List of square numbers.

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Random numbers, not square.

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Simplified numerical term of numerator.

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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 denominator.

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Random number.

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Multiple of LCM of n amd m.

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Random number.

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Simplified surd term of numerator.

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Random numbers, not square.

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GCD of terms in numerator.

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(a)

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

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i)

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

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ii)

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

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(b)

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

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

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

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

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

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

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So the whole expression is

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

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Finally, we can cancel down by noting that $\\sqrt{\\var{num}}=\\sqrt{\\var{num/(a[0]*b[0])} \\times \\var{a[0]*b[0]}} = \\sqrt{\\var{num/(a[0]*b[0])}} \\times \\sqrt{\\var{a[0]*b[0]}}=\\var{sqrt(num/(a[0]*b[0]))} \\times \\sqrt{\\var{a[0]*b[0]}}$

\n

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Hence we can simplify our fraction by cancelling down as follows:

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\\[ \\frac{\\sqrt{\\var{num}}}{\\var{m_lcm}} = \\frac{\\var{sqrt(num/(a[0]*b[0]))} \\times \\sqrt{\\var{a[0]*b[0]}}}{\\var{m_lcm}} = \\frac{\\sqrt{\\var{a[0]*b[0]}}}{\\var{b[0]}} \\]

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\n

(c)

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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 $\\var{d}\\times\\var{f}=\\var{df}$.

\n

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

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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

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

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

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\n

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(d)

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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.

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

\n

\\[ \\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{.} \\]

\n

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

\n

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

\n

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

\n

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

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which simplifies to

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

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\n

\n

(e)

\n

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

\n

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. 

\n

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

\n

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

\n

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

\n

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

\n

\n

\n

\n

(f)

\n

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.

\n

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 

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

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This can be simplified further by collecting like terms to obtain

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

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

\n

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. 

\n

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

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

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\\[ \\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}}} \\]

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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:

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

\n

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

\n

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

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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.

\n

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

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

\n

i)

\n

\n

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

\n

\n

ii)

\n

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

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

\n

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

\n

where

\n

$a =$ [[0]]

\n

$b =$ [[1]]

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

\n

\n

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

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

", "

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

", "

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

", "

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

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

\n

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

\n

where:

\n

$a = $ [[0]]

\n

$b = $ [[1]]

\n

$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.

\n

\\[ \\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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