# How do you simplify the square root of -72 + square root of -50?

Sep 13, 2015

$\sqrt{- 72} + \sqrt{- 50} = 11 i \sqrt{2}$

#### Explanation:

If $a , b \ge 0$ then $\sqrt{a b} = \sqrt{a} \sqrt{b}$

If $a < 0$ then $\sqrt{a} = i \sqrt{- a}$

So:

$\sqrt{- 72} = i \sqrt{72} = i \sqrt{{6}^{2} \cdot 2} = i \sqrt{{6}^{2}} \sqrt{2} = 6 i \sqrt{2}$

$\sqrt{- 50} = i \sqrt{50} = i \sqrt{{5}^{2} \cdot 2} = 1 \sqrt{{5}^{2}} \sqrt{2} = 5 i \sqrt{2}$

So:

$\sqrt{- 72} + \sqrt{- 50} = 6 i \sqrt{2} + 5 i \sqrt{2} = 11 i \sqrt{2}$

Note that the identity $\sqrt{a b} = \sqrt{a} \sqrt{b}$ does not hold if $a , b < 0$.

For example:

$1 = \sqrt{1} = \sqrt{- 1 \cdot - 1} \ne \sqrt{- 1} \cdot \sqrt{- 1} = - 1$