# How do you solve roots and radical questions that have fractions underneath the radical sign? For example the following question: 1/2√4/3 - 2/3√3/4 + 2√1/12

Nov 2, 2014

Let us look at the posted question step by step,

$\frac{1}{2} \sqrt{\frac{4}{3}} + \frac{2}{3} \sqrt{\frac{3}{4}} + 2 \sqrt{\frac{1}{12}}$

by distributing the square-roots to the numerator and the denominator,

$= \frac{1}{2} \frac{\sqrt{4}}{\sqrt{3}} + \frac{2}{3} \frac{\sqrt{3}}{\sqrt{4}} + 2 \frac{\sqrt{1}}{\sqrt{12}}$

by multiplying the numerator and the denominator by the denominator to rationalize the denominators,

$= \frac{1}{2} \frac{\sqrt{4} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} + \frac{2}{3} \frac{\sqrt{3} \cdot \sqrt{4}}{\sqrt{4} \cdot \sqrt{4}} + 2 \frac{\sqrt{1} \cdot \sqrt{12}}{\sqrt{12} \cdot \sqrt{12}}$

by simplifying a bit further,

$= \frac{1}{6} \sqrt{12} + \frac{1}{6} \sqrt{12} + \frac{1}{6} \sqrt{12}$

by $\sqrt{12} = 2 \sqrt{3}$,

$= \setminus \frac{1}{3} \sqrt{3} + \setminus \frac{1}{3} \sqrt{3} + \setminus \frac{1}{3} \sqrt{3}$

by factoring out $\sqrt{3}$,

$= \left(\frac{1}{3} + \frac{1}{3} + \frac{1}{3}\right) \sqrt{3} = \sqrt{3}$

I hope that this was helpful.