Simplify the expression sqrt -9/ (3-2i)+(1+5i)
first combine like terms
\[\huge\rm \frac{ \sqrt{-9} }{ 3-2i+1+5i }\]
12-9i/25 12-9i/7 9+12i/25 9+12i/7
and yes
square root of 9 over 4 plus 3i?
-9*
alright and you separate -9 \[\sqrt{-1} \times \sqrt{9}\] and i =what ?
yep right
i = imaginary number? or -1
i imaginary
general question
ok
\[\bf i = \sqrt{-1}\] \[\bf i^2 = -1\] and that's why we have to seprate of negative one and positive 9
ok
\[\huge\rm \frac{ \sqrt{-1} \times \sqrt{9} }{ 4-3i }\] so replace sqrt -1 by 1 and multiply both the denominator and the numerator by the conjugate of 4-3i
-9/25+12i/25?
nope how did you get 25
just guessed
why did you guess ?
you are not 'n the exam haall this is the time to learn not to guess :=)
Honestly I have no idea how to do this and not trying to be rude but i need the answer because i am taking a timed exam.......
really so it's a test question ..sorryy..
yes
do you know we are not allowed to help u on your test or quiz? it's against the coc
I know but i really dont understand this and i have also seen alot of people do the same thing as me
well..if someone don't aa bad thing then.. you don't have to follow them you will do better without cheating..
ok, thanks for the help anway
what is square root of 9? replace square root of -1 by i multiply by the conjugate 4-3i if you don't know what conjugate then search it
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