# How do you simplify 2sqrt(-49) + 3sqrt(-64)?

Nov 17, 2015

$= \pm 38 i \mathmr{and} \pm 10 i$

#### Explanation:

From complex number theory, $i = \sqrt{- 1} \mathmr{and} {i}^{2} = - 1$.

Note also that ${\left(a i\right)}^{2} = {\left(- a i\right)}^{2} = - 1$ and hence $\sqrt{- a} = \pm i \sqrt{a}$ , $\forall a \in {\mathbb{R}}^{+}$.

$\therefore 2 \sqrt{- 49} + 3 \sqrt{- 64} = 2 \cdot \left(\pm 7 i\right) + 3 \cdot \left(\pm 8 i\right)$

$= \pm 14 i \pm 24 i$

$= \pm 38 i \mathmr{and} \pm 10 i$

Nov 17, 2015

$34 i$

#### Explanation:

Remember that ${i}^{2} = 1$ so $\sqrt{- 1} = i$.

Keeping this in mind, you can simplify the term as follows:

$\textcolor{w h i t e}{\times} 2 \sqrt{- 49} + 3 \sqrt{- 64}$

$= 2 \cdot \sqrt{- 1} \cdot \sqrt{49} + 3 \cdot \sqrt{- 1} \cdot \sqrt{64}$

$= 2 \cdot i \cdot 7 + 3 \cdot i \cdot 8$

$= 14 i + 24 i$

$= 34 i$