# How do you subtract this and answer using standard form of a complex number(a+bi)?

## $\left(3 + 2 \sqrt{54}\right) - \left(- 2 - \sqrt{- 24}\right)$

Dec 20, 2016

$\textcolor{b l u e}{5 + 6 \sqrt{6}} + \textcolor{g r e e n}{2 \sqrt{6}} \textcolor{red}{i}$

#### Explanation:

The standard form is $\textcolor{b l u e}{a} + \textcolor{g r e e n}{b} \textcolor{red}{i}$

$\left(3 + 2 \sqrt{54}\right) - \left(- 2 - \sqrt{- 24}\right)$

Distribute the negative

$= \left(3 + 2 \sqrt{54}\right) + 2 + \sqrt{- 24}$

$= 3 + 2 \sqrt{54} + 2 + \sqrt{- 24}$

$= 5 + 2 \sqrt{54} + \sqrt{- 24}$

$\textcolor{b r o w n}{\sqrt{- 24} = i \sqrt{24}}$

$= 5 + 2 \sqrt{54} + i \sqrt{24}$

You can stop here and your answer would be $\textcolor{b l u e}{5 + 2 \sqrt{54}} + \textcolor{g r e e n}{\sqrt{24}} \textcolor{red}{i}$, or you can simplify the radicals (optional)

$5 + 2 \sqrt{54} + i \sqrt{24}$

$= 5 + 2 \sqrt{9 \cdot 6} + i \sqrt{4 \cdot 6}$

$= 5 + 2 \sqrt{9} \sqrt{6} + i \sqrt{4} \sqrt{6}$

$= 5 + 2 \cdot 3 \sqrt{6} + i \cdot 2 \sqrt{6}$

$= 5 + 6 \sqrt{6} + 2 i \sqrt{6}$

$= 5 + 6 \sqrt{6} + 2 \sqrt{6} i$

and your answer in the standard form would be $\textcolor{b l u e}{5 + 6 \sqrt{6}} + \textcolor{g r e e n}{2 \sqrt{6}} \textcolor{red}{i}$