# How do you evaluate square root of 4/3 minus the square root of 3/4?

Apr 9, 2018

#### Explanation:

Let $a = \sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$
Let $b = \sqrt{\frac{4}{3}} + \sqrt{\frac{3}{4}}$

$a b = \frac{4}{3} - \frac{3}{4}$
$a b = \frac{7}{12}$
$a + b = 2 \sqrt{\frac{4}{3}}$
$a + b = \frac{4}{\sqrt{3}}$

Now as simultaneous equations:
$a \left(\frac{4}{\sqrt{3}} - a\right) = \frac{7}{12}$
${a}^{2} - \frac{4 a}{\sqrt{3}} + \frac{7}{12} = 0$

$a = \frac{\frac{4}{\sqrt{3}} \pm \sqrt{{\left(\frac{4}{\sqrt{3}}\right)}^{2} - 4 \times \frac{7}{12}}}{2}$

$a = \frac{\frac{4}{\sqrt{3}} \pm \sqrt{\frac{16}{3} - \frac{7}{3}}}{2}$

$a = \frac{\frac{4}{\sqrt{3}} \pm \frac{3}{\sqrt{3}}}{2}$

$a = \frac{\frac{4}{\sqrt{3}} - \frac{3}{\sqrt{3}}}{2}$

$a = \frac{1}{2 \sqrt{3}}$

Therefore $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}} = \frac{1}{2 \sqrt{3}}$
and we also get $\sqrt{\frac{4}{3}} + \sqrt{\frac{3}{4}} = \frac{7}{2 \sqrt{3}}$