# How to add and subtract radical equations (3sqrt2)/9 +1/(2sqrt32)+sqrt9/8?

Apr 3, 2015

We first simplify the individual terms as much as possible

$\frac{3 \sqrt{2}}{9} = \frac{\sqrt{2}}{3}$ (divide upper and lower by $3$)

$32 = 2 \cdot 16 = 2 \cdot {4}^{2} \to \sqrt{32} = 4 \sqrt{2} \to$
1/(2sqrt32)=1/(8sqrt2

$9 = {3}^{2} \to \frac{\sqrt{9}}{8} = \frac{3}{8}$

Our problem has now evolved into:

$\frac{\sqrt{2}}{3} + \frac{1}{8 \sqrt{2}} + \frac{3}{8}$

We need to take care of the second term by multiplying (above and below) by $\sqrt{2}$

$\frac{1}{8 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{8 \cdot 2} = \frac{\sqrt{2}}{16}$

Now we need to get all numerators alike; we'll have to get them to $48$ (= smallest common multiple)

$\frac{16 \cdot \sqrt{2}}{16 \cdot 3} + \frac{3 \cdot \sqrt{2}}{3 \cdot 16} + \frac{6 \cdot 3}{6 \cdot 8} =$

$\frac{16 \sqrt{2} + 3 \sqrt{2} + 18}{48} = \frac{19 \sqrt{2} + 18}{48} = \frac{19}{48} \sqrt{2} + \frac{3}{8}$