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The best approach here is to first rationalise the denominator. We do this by multiplying both top and bottom by an appropriate value.

For example to rationalise the denominator for an expression like $\\frac{a}{\\sqrt{b}}$, we multiply numerator and denominator by $\\sqrt{b}$ to get $\\frac{a\\sqrt{b}}{b}$

Similarly to rationalise the denominator for an expression like $\\frac{a+\\sqrt{b}}{c+\\sqrt{d}}$, we multiply numerator and denominator by $c-\\sqrt{d}$ to get $\\frac{(a+\\sqrt{b})(c-\\sqrt{d})}{c^2-d}$

Dont forget that:

$\\sqrt{a \\times b} = \\sqrt{a} \\times \\sqrt{b}$

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

$(a+\\sqrt{b})(a-\\sqrt{b})=a^2-b$

Remember to check that your answer is in its simplest form.

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1$\\var{a1}-\\sqrt{\\var{a2}}$ $=$ [[0]]+$\\sqrt{\\var{a2}}$[[1]]

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In this question we find an irrational surd in the denominator. This time, however, it is one of two terms which make up a binomial expression in the denominator.

Rationalising the denominator cannot simply be done by multiplying the numerator and denominator by the surd, as there will still be a surd remaining in the denominator:

$\\frac{1}{\\var{a1}-\\sqrt{\\var{a2}}}\\times\\frac{\\sqrt{\\var{a2}}}{\\sqrt{\\var{a2}}}=\\frac{\\sqrt{\\var{a2}}}{\\var{a1}\\sqrt{\\var{a2}}-\\var{a2}}$

Instead, we take the conjugate of the binomial expression in the denominator. This is essentially the same expression but with the sign of the second term changed.

i.e. the conjugate of $\\var{a1}-\\sqrt{\\var{a2}}$ is $\\var{a1}+\\sqrt{\\var{a2}}$.

We then multiply the numerator and denominator by this conjugate and simplify:

$\\frac{1}{\\var{a1}-\\sqrt{\\var{a2}}}\\times\\frac{\\var{a1}+\\sqrt{\\var{a2}}}{\\var{a1}+\\sqrt{\\var{a2}}}=\\frac{\\var{a1}+\\sqrt{\\var{a2}}}{(\\var{a1}-\\sqrt{\\var{a2}})(\\var{a1}+\\sqrt{\\var{a2}})}$.

You may now recognise the pattern of binomial expression multiplied by its conjugate as the factorised equivalent of the difference of two squares, e.g. $(x+2)(x-2)=x^2-4$.

Using this principle, the denominator simplifies as follows:

$\\frac{\\var{a1}+\\sqrt{\\var{a2}}}{(\\var{a1}-\\sqrt{\\var{a2}})(\\var{a1}+\\sqrt{\\var{a2}})}=\\frac{\\var{a1}+\\sqrt{\\var{a2}}}{(\\var{a1})^2-\\var{a2}}=\\frac{\\var{a1}+\\sqrt{\\var{a2}}}{\\simplify{{a1}^2-{a2}}}$

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1$\\var{b1}-\\sqrt{\\var{b2}}$ $=$ [[0]]+$\\sqrt{\\var{b2}}$[[1]]

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Again, we multiply the numerator and denominator by the denominator conjugate and simplify:

$\\frac{1}{\\var{b1}-\\sqrt{\\var{b2}}}\\times\\frac{\\var{b1}+\\sqrt{\\var{b2}}}{\\var{b1}+\\sqrt{\\var{b2}}}=\\frac{\\var{b1}+\\sqrt{\\var{b2}}}{(\\var{b1}-\\sqrt{\\var{b2}})(\\var{b1}+\\sqrt{\\var{b2}})}=\\frac{\\var{b1}+\\sqrt{\\var{b2}}}{(\\var{b1})^2-\\var{b2}}=\\frac{\\var{b1}+\\sqrt{\\var{b2}}}{\\simplify{{b1}^2-{b2}}}$.

$\\var{c3}$$\\var{c1}-\\sqrt{\\var{c2}}$$=$[[0]]$+$[[1]]$\\sqrt{\\var{c2}}$[[2]]

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Again, we multiply the numerator and denominator by the denominator conjugate and simplify:

$\\frac{\\var{c3}}{\\var{c1}-\\sqrt{\\var{c2}}}\\times\\frac{\\var{c1}+\\sqrt{\\var{c2}}}{\\var{c1}+\\sqrt{\\var{c2}}}=\\frac{\\var{c3}\\big(\\var{c1}+\\sqrt{\\var{c2}}\\big)}{(\\var{c1}-\\sqrt{\\var{c2}})(\\var{c1}+\\sqrt{\\var{c2}})}=\\frac{\\var{c3}\\big(\\var{c1}+\\sqrt{\\var{c2}}\\big)}{\\big((\\var{c1})^2-\\var{c2}\\big)}=\\frac{\\big(\\var{c6}\\times\\var{c1}\\big)+\\big(\\var{c6}\\times\\sqrt{\\var{c2}}\\big)}{\\var{c7}}=\\frac{\\simplify{{c6}*{c1}}+\\var{c6}\\sqrt{\\var{c2}}}{\\var{c7}}$

$\\var{d3}$$\\var{d1}-\\sqrt{\\var{d2}}$ $=$ [[0]]$+$[[1]]$\\sqrt{\\var{d2}}$[[2]]

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Again, we multiply the numerator and denominator by the denominator conjugate and simplify:

$\\frac{\\var{d3}}{\\var{d1}-\\sqrt{\\var{d2}}}\\times\\frac{\\var{d1}+\\sqrt{\\var{d2}}}{\\var{d1}+\\sqrt{\\var{d2}}}=\\frac{\\var{d3}\\big(\\var{d1}+\\sqrt{\\var{d2}}\\big)}{(\\var{d1}-\\sqrt{\\var{d2}})(\\var{d1}+\\sqrt{\\var{d2}})}=\\frac{\\var{d3}\\big(\\var{d1}+\\sqrt{\\var{d2}}\\big)}{\\big((\\var{d1})^2-\\var{d2}\\big)}=\\frac{\\var{d3}\\big(\\var{d1}+\\sqrt{\\var{d2}}\\big)}{\\var{d4}}=\\frac{\\big(\\var{d6}\\times\\var{d1}\\big)+\\big(\\var{d6}\\times\\sqrt{\\var{d2}}\\big)}{\\var{d7}}=\\frac{\\simplify{{d6}*{d1}}+\\var{d6}\\sqrt{\\var{d2}}}{\\var{d7}}$

$\\var{e1}-\\sqrt{\\var{e3}}$$\\var{e2}-\\sqrt{\\var{e3}}$ $=$ [[0]]$+$[[1]]$\\sqrt{\\var{e3}}$[[2]]

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Again, we multiply the numerator and denominator by the denominator conjugate and simplify:

$\\frac{\\var{e1}-\\sqrt{\\var{e3}}}{\\var{e2}-\\sqrt{\\var{e3}}}\\times\\frac{\\var{e2}+\\sqrt{\\var{e3}}}{\\var{e2}+\\sqrt{\\var{e3}}}=\\frac{(\\var{e1}-\\sqrt{\\var{e3}})(\\var{e2}+\\sqrt{\\var{e3}})}{(\\var{e2}-\\sqrt{\\var{e3}})(\\var{e2}+\\sqrt{\\var{e3}})}=\\frac{\\big(\\var{e1}\\times\\var{e2}\\big)+\\var{e1}\\sqrt{\\var{e3}}-\\var{e2}\\sqrt{\\var{e3}}-\\var{e3}}{\\big((\\var{e2})^2-\\var{e3}\\big)}=\\frac{\\big(\\var{e1}\\times\\var{e2}\\big)+\\big(\\var{e1}-\\var{e2}\\big)\\sqrt{\\var{e3}}-\\var{e3}}{\\big((\\var{e2})^2-\\var{e3}\\big)}=\\frac{\\var{e4}+\\var{e5}\\sqrt{\\var{e3}}}{\\var{e6}}=\\frac{\\simplify{{e9}+{e10}*sqrt{e3}}}{\\var{e11}}$

Note: The coefficient of the surd in the result may be negative, but the question requires you type the answer into the box next to a plus sign. If it is indeed negative, write the negative number but keep in mind that the two signs would simplify to a negative when written.

$\\var{f1}+\\sqrt{\\var{f3}}$$\\var{f2}-\\sqrt{\\var{f3}}$$ = $ [[0]]$+$[[1]]$\\sqrt{\\var{f3}}$[[2]]

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Again, we multiply the numerator and denominator by the denominator conjugate and simplify:

$\\frac{\\var{f1}+\\sqrt{\\var{f3}}}{\\var{f2}-\\sqrt{\\var{f3}}}\\times\\frac{\\var{f2}+\\sqrt{\\var{f3}}}{\\var{f2}+\\sqrt{\\var{f3}}}=\\frac{(\\var{f1}+\\sqrt{\\var{f3}})(\\var{f2}+\\sqrt{\\var{f3}})}{(\\var{f2}-\\sqrt{\\var{f3}})(\\var{f2}+\\sqrt{\\var{f3}})}=\\frac{\\big(\\var{f1}\\times\\var{f2}\\big)+\\var{f1}\\sqrt{\\var{f3}}+\\var{f2}\\sqrt{\\var{f3}}+\\var{f3}}{\\big((\\var{f2})^2-\\var{f3}\\big)}=\\frac{\\big(\\var{f1}\\times\\var{f2}\\big)+\\big(\\var{f1}+\\var{f2}\\big)\\sqrt{\\var{f3}}+\\var{f3}}{\\big((\\var{f2})^2-\\var{f3}\\big)}=\\frac{\\var{f4}+\\var{f5}\\sqrt{\\var{f3}}}{\\var{f6}}=\\frac{\\simplify{{f9}+{f10}*sqrt{f3}}}{\\var{f11}}$

$\\var{g1}+\\sqrt{\\var{g13}}$$\\var{g2}+2\\sqrt{\\var{g3}}$$ =$ [[0]]$+$[[1]]$\\sqrt{\\var{g3}}$[[2]]

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Again, we multiply the numerator and denominator by the denominator conjugate and simplify:

$\\frac{\\var{g1}+\\sqrt{\\var{g13}}}{\\var{g2}+2\\sqrt{\\var{g3}}}\\times\\frac{\\var{g2}-2\\sqrt{\\var{g3}}}{\\var{g2}-2\\sqrt{\\var{g3}}}=\\frac{(\\var{g1}+\\sqrt{\\var{g13}})(\\var{g2}-2\\sqrt{\\var{g3}})}{(\\var{g2}+2\\sqrt{\\var{g3}})(\\var{g2}-2\\sqrt{\\var{g3}})}=\\frac{\\big(\\var{g1}\\times\\var{g2}\\big)+\\var{g2}\\sqrt{\\var{g13}}-\\simplify{2*{g1}}\\sqrt{\\var{g3}}-2\\sqrt{\\var{g3}\\times\\var{g13}}}{\\big((\\var{g2})^2-4\\var{g3}\\big)}=\\frac{\\simplify{{g1}*{g2}}+\\var{g2}\\sqrt{\\var{g3}\\times\\var{g12}}-\\simplify{2*{g1}}\\sqrt{\\var{g3}}-2\\sqrt{\\var{g3}\\times\\var{g3}\\times\\var{g12}}}{\\simplify{{g2}^2}-4\\var{g3}}=\\frac{\\var{g4}+\\sqrt{\\var{g3}}\\big(\\var{g2}\\sqrt{\\var{g12}}-\\simplify{2*{g1}}\\big)}{\\var{g6}}=\\frac{\\var{g4}+\\var{g5}\\sqrt{\\var{g3}}}{\\var{g6}}=\\frac{\\simplify{{g9}+{g10}*sqrt{g3}}}{\\var{g11}}$

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'Rationalise the denominators' of the following expressions and present each in the form $\\frac{a+b\\sqrt{c}}{d}$ where a, b, c and d are integers, with a, b, and d having no common factors.

With simpler operations, it's easier to manipulate fractions which contain surds in the numerator instead of the denominator. The process of transferring the surd from the denominator to the numerator is called 'ratinoalising the denominator'. If you've never done a question like this before, click on 'Show steps' and follow the solutions, or click 'Reveal answers' at the bottom of the page and study the Advice section. Once you are confident with the method, click 'Try another question like this one' and complete the questions without clicking for help.

Note: the final answer may simplify in such a way that the number $1$ is in the place of numerator coefficient, in the denominator or both. In these cases the $1$ wouldn't be written in the final answer but this question requires that you write it into the given gaps.

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e3", "templateType": "anything", "group": "Ungrouped variables", "name": "e6", "description": ""}, "c7": {"definition": "c4/c5", "templateType": "anything", "group": "Ungrouped variables", "name": "c7", "description": ""}, "c6": {"definition": "c3/c5", "templateType": "anything", "group": "Ungrouped variables", "name": "c6", "description": ""}, "c5": {"definition": "gcd(c3,c4)", "templateType": "anything", "group": "Ungrouped variables", "name": "c5", "description": ""}, "c4": {"definition": "c1^2 - c2", "templateType": "anything", "group": "Ungrouped variables", "name": "c4", "description": ""}, "e7": {"definition": "gcd(e4,e5)", "templateType": "anything", "group": "Ungrouped variables", "name": "e7", "description": ""}, "e10": {"definition": "e5/e8", "templateType": "anything", "group": "Ungrouped variables", "name": "e10", "description": ""}, "f10": {"definition": "f5/f8", "templateType": "anything", "group": "Ungrouped variables", "name": "f10", "description": ""}, "f11": {"definition": "f6/f8", "templateType": "anything", "group": "Ungrouped variables", "name": "f11", "description": ""}, "g13": {"definition": "g3*g12", "templateType": "anything", "group": "Ungrouped variables", "name": "g13", "description": ""}, "g12": {"definition": "random(4,9,16,25)", "templateType": "anything", "group": "Ungrouped variables", "name": "g12", "description": ""}, "g11": {"definition": "g6/g8", "templateType": "anything", "group": "Ungrouped variables", "name": "g11", "description": ""}, "g10": {"definition": "g5/g8", "templateType": "anything", "group": "Ungrouped variables", "name": "g10", "description": ""}}, "metadata": {"description": "

Practice rationalising a binomial denominator which contains surds.

Adapted from 'Surds 3' by Joshua Boddy.

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