What is the 3square root of 50 - square root of 32 + 6 square root of 128?

Sep 8, 2015

$3 \sqrt{50} - \sqrt{32} + 6 \sqrt{128} = 59 \sqrt{2}$

Explanation:

Use $\sqrt{a b} = \sqrt{a} \sqrt{b}$ and $\sqrt{{a}^{2}} = a$ if $a \ge 0$

$3 \sqrt{50} - \sqrt{32} + 6 \sqrt{128} =$

$3 \sqrt{{5}^{2} \cdot 2} - \sqrt{{4}^{2} \cdot 2} + 6 \sqrt{{8}^{2} \cdot 2}$

$3 \sqrt{{5}^{2}} \sqrt{2} - \sqrt{{4}^{2}} \sqrt{2} + 6 \sqrt{{8}^{2}} \sqrt{2}$

$= \left(3 \cdot 5\right) \sqrt{2} - 4 \sqrt{2} + \left(6 \cdot 8\right) \sqrt{2}$

$= \left(15 - 4 + 48\right) \sqrt{2}$

$= 59 \sqrt{2}$