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Given the fraction \\[\\dfrac{\\var{questnum}}{\\sqrt{\\var{surd1}}\\var{densign}\\sqrt{\\var{surd2}}}\\] we can rationalise the denominator and rewrite the fraction in the simplified equivalent form.

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[[0]]    $\\,\\,\\,\\,$[[1]]$+$ [[2]]    $\\,\\,\\,\\,$[[3]]
[[4]]
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Note: This question requires the larger surd term to be entered on the left and the smaller one on the right e.g. $\\frac{2\\sqrt{7} + 5\\sqrt{3}}{6}$.

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In the above case, to rationalise the denominator we can multiply the top and bottom of the fraction by the conjugate surd of the denominator. This will rationalise the denominator since $\\left(\\sqrt{a}+\\sqrt{b}\\right)\\times\\left(\\sqrt{a}-\\sqrt{b}\\right)=a-\\sqrt{ab}+\\sqrt{ab}-b=a-b$. 

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$\\dfrac{\\var{questnum}}{\\sqrt{\\var{surd1}}\\var{densign}\\sqrt{\\var{surd2}}}$$=$$\\dfrac{\\var{questnum}}{\\sqrt{\\var{surd1}}\\var{densign}\\sqrt{\\var{surd2}}}\\times\\dfrac{\\sqrt{\\var{surd1}}\\var{conjsign}\\sqrt{\\var{surd2}}}{\\sqrt{\\var{surd1}}\\var{conjsign}\\sqrt{\\var{surd2}}}$    (multiplying top and bottom by the conjugate surd of the denominator)
$=$$\\dfrac{\\var{questnum}\\sqrt{\\var{surd1}}\\var{conjsign}\\var{questnum}\\sqrt{\\var{surd2}}}{\\var{surd1}-\\var{surd2}}$    
$=$$\\dfrac{\\var{questnum}\\sqrt{\\var{surd1}}\\var{conjsign}\\var{questnum}\\sqrt{\\var{surd2}}}{\\var{tempden}}$    
$=$$\\dfrac{\\var{ansnummult}\\sqrt{\\var{surd1}}\\var{conjsign}\\var{ansnummult}\\sqrt{\\var{surd2}}}{\\var{ansden}}$    (cancelling the common factor of $\\var{cancel}$)
$=$$\\var{ansnummult}\\sqrt{\\var{surd1}}\\var{conjsign}\\var{ansnummult}\\sqrt{\\var{surd2}}$    (cancelling the common factor of $\\var{cancel}$)
$=$$\\var{ansnummult}\\sqrt{\\var{surd1}}\\var{conjsign}\\var{ansnummult}\\sqrt{\\var{surd2}}$
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Single example with detailed solution for rationalising a binomial denominator which contains surds.

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Adapted from 'Surds: rationalising the denominator conjugate' by Ben Brawn.

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