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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 

\n
\n\n\n\n\n\n\n\n\n\n\n\n\n
[[0]]    $\\,\\,\\,\\,$[[1]]$+$ [[2]]    $\\,\\,\\,\\,$[[3]]
[[4]]
\n

\n

Note: If you are marked incorrect, please try swapping the order of terms in the numerator, this question requires the larger surd part on the left and the smaller surd part on the right. 

\n
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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 $\\left(\\sqrt{a}+\\sqrt{b}\\right)\\times\\left(\\sqrt{a}-\\sqrt{b}\\right)=a-\\sqrt{ab}+\\sqrt{ab}-b=a-b$. 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\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}}$
\n

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