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My teacher gave this challenger in class today and I can't figure how to solve it: $\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$

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    For your title question: $\dfrac{\sqrt{\dfrac{a}{b}}}{\sqrt{\dfrac{c}{d}}}={\sqrt{\dfrac{a}{b}}}\times{\sqrt{\dfrac{d}{c}}}={\sqrt{\dfrac{ad}{bc}}}$ but that is not really the issue with your teacher's challenge – 2017-01-20

2 Answers 2

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Given,

$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$

On rationalizing,

$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}} \cdot \frac{\sqrt{\frac13} + \sqrt{\frac12}}{\sqrt{\frac13} + \sqrt{\frac12}}$

= $\frac{\sqrt{\frac29} + \sqrt{\frac13} - \sqrt{\frac12} - \sqrt{\frac34}}{\frac{-1}{6}}$

= $-6 \left( \frac{\sqrt2}{3} + \frac{1}{\sqrt 3} - \frac{1}{\sqrt 2} - \frac{\sqrt3}{2} \right)$

= $-6 \left( \frac{2\sqrt2 + 2\sqrt3 - 3\sqrt2 - 3\sqrt3}{6} \right)$

= $- \left( - \sqrt2 - \sqrt3 \right)$

= $\sqrt2 + \sqrt3$

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    If any doubt please ask. – 2017-01-20
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    I did the rationalizing and got: (sqrt(2/9) + sqrt(1/3) - sqrt(1/2) - sqrt(3/4)) / (-1/6) – 2017-01-20
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    @uema: It might then be worth multiplying numerator and denominator by $-6$ – 2017-01-20
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    @Henry Ok, now I have: -6(sqrt(2/9) + sqrt(1/3) - sqrt(1/2) - sqrt(3/4)) – 2017-01-20
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    I think I solved it, is it sqrt(2) + sqrt(3)? – 2017-01-20
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    Yes you did it. Nice. I also solved it for you. – 2017-01-20
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    Any doubt you have? – 2017-01-20
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    @KanwaljitSingh Nope, just that. Thank you two! – 2017-01-21
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mutiplying numerator and denominator by $\sqrt{6}$ gives: $$ \frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}} = \frac{2-3}{\sqrt{2}-\sqrt{3}} = \frac1{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2} $$