Question

Rationalize the denominator of square root of -36 over (2 - 3i) + (3 + 2i).

Answer 1

I misread the q.

Answer 2

as did I haha

Answer 3

\[ \frac{\sqrt{ -36}}{(2 - 3i) + (3 + 2i)} = \frac{\sqrt{ -1 \times 36}}{2 - 3i + 3 + 2i}\]
\[= \frac{6\sqrt{ -1 }}{5 - i }\]
\[= \frac{6i}{5 - i }= \frac{(6i) \times (5+i)}{(5 - i) (5+i) }\]
\[= \frac{30i+6i^2}{(5^2 - i^2) }= \frac{30i+6(-1))}{25 -(-1) }\]
\[i = \sqrt {-1} \rightarrow i^2=-1 \]
\[= \frac{30i-6}{25 +1 }= \frac{-6+30i}{26 }\] is the rationalized form of the given expression.

@brooklynransome

Answer 4

\[\frac{-6+30i}{26} = \frac{2(-3+15i)}{26}= \frac{-3+15i}{13}\]

@brooklynransome