Rationalise the denominator with increasingly difficult examples involving compound denominators.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Rationalising the denominator means removing any surds on the denominator.
\nTry the following questions to practise rationalising with expressions in different forms.
", "advice": "The rule that should be used is $\\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}} = \\sqrt{\\frac{a}{b}}$.
\ni)
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{{\\var{prime_nums[0]}}}} = \\sqrt{\\var{square_nums[0]}}= \\simplify{{sqrt(square_nums[0])}}$; and
\nii)
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{{\\var{prime_nums[1]}}}} = \\sqrt{\\var{square_nums[1]}} = \\simplify{{sqrt(square_nums[1])}}$.
\nTo rationalise the denominator, you have to use the rule $\\displaystyle\\sqrt{a}\\times\\sqrt{a}=a$.
\nYou are asked to rationalise the denominator of the expression $\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}$.
\nAs ${\\sqrt{\\var{m_lcm}}}$ is the denominator, we must multiply the whole fraction by ${\\sqrt{\\var{m_lcm}}}$.
\n\\[ \\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}\\times\\frac{\\sqrt{\\var{m_lcm}}}{\\sqrt{\\var{m_lcm}}} \\]
\nBy using the rule $\\sqrt{a}\\times\\sqrt{b}$ = $\\sqrt{ab}$, we can get the numerator as $\\sqrt{\\var{num}}$.
\nThe denominator is ${\\sqrt{\\var{m_lcm}}}\\times{\\sqrt{\\var{m_lcm}}} = {\\var{m_lcm}}$.
\nSo the whole expression is
\n\\[ \\frac{\\sqrt{\\var{num}}}{\\var{m_lcm}} \\]
\nFinally, we can cancel down by finding the greatest common divisor of the top and bottom to obtain
\n\\[ \\frac{\\sqrt{\\var{a[0]*b[0]}}}{\\var{b[0]}} \\]
\nThe 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$\\displaystyle\\frac{\\var{c}}{\\var{d}\\sqrt{\\var{f}}}\\times\\simplify[all,!simplifyFractions,!sqrtDivision]{sqrt({f})/(sqrt({f}))}={\\frac{\\simplify{{c}*sqrt({f})}}{{(\\var{d}\\times\\var{f})}}}=\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}$.
\nMake 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
\n\\[\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}=\\frac{\\simplify{{c_coprime}*sqrt({f})}}{\\var{df_coprime}}\\]
\nWe are asked to find an answer in the form $\\frac{a \\sqrt{\\var{f}}}{b}$, so $a = \\var{c_coprime}$ and $b = \\var{df_coprime}$.
\n\nWhen 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.
\nThe 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{.} \\]
\nFrom 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$\\displaystyle\\frac{\\var{g}(\\sqrt{\\var{h}}-\\var{j})}{(\\sqrt{\\var{h}}+\\var{j})(\\sqrt{\\var{h}}-\\var{j})}==\\frac{\\simplify[all,!collectNumbers,!noLeadingMinus]{({-{g}*{j}})+{g}}\\sqrt{\\var{h}}}{\\var{h}{-\\simplify{{j}*sqrt({h})}}+\\simplify{{j}*sqrt({h})}-{{\\var{j^2}}}}$.
\nThe 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{,} \\]
\nwhich simplifies to
\n\\[ \\var{o}\\sqrt{\\var{h}}-{\\var{l}}\\text{.} \\]
\n\n$\\displaystyle\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}}$ =
\nTo 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$\\displaystyle\\frac{\\var{t^2}-\\var{2t}\\sqrt{\\var{u}}+\\var{u}}{\\var{t^2}-\\var{u}}=\\frac{\\var{t^2+u}-\\var{2t}{\\sqrt{\\var{u}}}}{\\var{t^2-u}}$.
\nDividing 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\nTo 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.
\nIt 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
\n\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\var{p}\\sqrt{\\var{p}}} \\text{.} \\]
\nThis can be simplified further by collecting like terms to obtain
\n\\[ \\frac{\\var{p}}{\\var{p+1}\\sqrt{\\var{p}}} \\text{.} \\]
\nTherefore, the expression is now $\\displaystyle\\frac{\\var{p}}{{\\var{p+1}}\\sqrt{\\var{p}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}}$.
\nIn 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$\\displaystyle\\frac{\\var{r_coprime}}{\\var{s_coprime}}\\times\\simplify[all,!simplifyFractions,!sqrtDivision]{{p_p1}*sqrt({p})/({p_p1}*sqrt({p}))}=\\frac{\\simplify{{one}*sqrt({p})}}{\\var{two}\\sqrt{\\var{p}}}$,
\nBoth parts of the expression now have the same denominator and we can add them.
\n\\[ \\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}}} \\]
\nWhen 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:
\n\\[ \\frac{\\sqrt{\\var{p}}(\\sqrt{\\var{p}}+\\var{r_coprime*p_p1})}{\\var{p+1}\\sqrt{\\var{p}}}\\text{.} \\]
\nWe 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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\ni)
\n\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}}=$ [[0]]
\n\nii)
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}}=$ [[1]]
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\n\n| $\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}=$ | \n$\\surd$[[0]] | \n
| [[1]] | \n
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{,} \\]
\nwhere
\n$a =$ [[0]]
\n$b =$ [[1]]
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\n\n$\\displaystyle\\frac{\\var{g}}{(\\sqrt{\\var{h}}+\\var{j})}$ =
", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["$\\displaystyle\\var{o}\\sqrt{\\var{h}}+\\var{l}$
", "$\\displaystyle-\\var{l}-\\var{o}\\sqrt{\\var{h}}$
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", "$\\displaystyle\\var{l}-\\sqrt{\\var{h}}$
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\n\\[ \\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}} = \\frac{a-b\\sqrt{\\var{u}}}{c} \\text{,} \\]
\nwhere:
\n$a = $ [[0]]
\n$b = $ [[1]]
\n$c = $ [[2]]
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\n\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\sqrt{\\var{p^3}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}} \\]
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