Question

help. will give medal. step by step plse. simplify the expression -5+i/2i

Answer 1

-5+i/2i=-5-i/2i
-5-i/2*i=-5/2=-2.5

Answer 2

Hey panda, is the 2i All in the bottom of this fraction?
It's hard to interpret the problem with the way you wrote it.

Like this \(\large\rm \frac{-5+i}{2i}\) or like this \(\large\rm -5+\frac{i}{2i}\)

Answer 3

whyd u subtract the i and why multiply 2*i?

Answer 4

@zepdrix yeah it the first fraction you wrote

Answer 5

To take the variable out

Answer 6

oh ok so -2.5 would be my answer?

Answer 7

I guess a good first step might be to multiply top and bottom by i,\[\large\rm \frac{-5+i}{2i}\cdot\color{royalblue}{\frac{i}{i}}\]That will turn your denominator into a real number, ya?

Answer 8

ok

Answer 9

\(\large\rm 2i\cdot i=2i^2=-2\)
That's how the denominator will clean up.
In the numerator, distribute the i to both terms,\[\large\rm \frac{-5i+i^2}{-2}\]

Answer 10

how di u get -2?

Answer 11

Yes... How did you? Cx

Answer 12

Recall that i is defined to be the square root of -1.\[\large\rm i=\sqrt{-1}\]So when we multiply i's together, they become a real number! :)
\[\large\rm i\cdot i\quad= \color{royalblue}{i}^2\quad= \color{royalblue}{(\sqrt{-1})}^2\quad= -1\]The square and the root undo one another.

Answer 13

oooh ok i get it

Answer 14

All I do is take out the variables. I did mess up when I wrote it, but the answer is the same when I do it. -5+i/2i=-5+i=-5-i=-5/2*i=2 So I get -5/2=-2.5... Can you help me figure out what I'm doing wrong?

Answer 15

ugh im confused......

Answer 16

Tubs, you can only take the i out of the second part, not out of the entire thing,
that's where the mistake is coming.

If we split up our fraction like this:\[\large\rm \frac{-5+i}{2i}\quad=\frac{-5}{2i}+\frac{i}{2i}\]NOW you're allowed to divide the i's out in the second fraction.
But not the way you did it before because the 5 didn't have an i.
At least I think that's what you did... I'm not sure.

Answer 17

I think multiplying by i is confusing you, let's try doing it this way instead Panda.
So you understand how I wrote it as two fractions with the same denominator? :o

Answer 18

yes

Answer 19

Oh! I get it! I should have divided by i. Sorry

Answer 20

In the second equation, we can divide an i out of the top and bottom.\[\large\rm \frac{\cancel{i}}{2\cancel{i}}\quad=\frac{1}{2}\]

Answer 21

in the second term*

Answer 22

I get it now. Sorry I messed you guys up! :/

Answer 23

ok im understanding so far

Answer 24

\[\large\rm =\frac{-5}{2i}+\frac{1}{2}\]In the first term,
here is a nice little trick you can remember:\[\large\rm \frac{1}{i}=-i\]

Answer 25

ok

Answer 26

So we'll bring the i up to the numerator, and apply a negative,\[\large\rm =-\frac{-5i}{2}+\frac{1}{2}\]

Answer 27

s0 -5i-1?

Answer 28

? 0_o

Answer 29

The negative is being applied to the first term.
But notice it already has a negative, ya?
So those two negatives will multiply to become positive.\[\large\rm =\frac{5i}{2}+\frac{1}{2}\]

Answer 30

so -5-1= 5i?

Answer 31

I don't understand what you're doing :( Hmm

Answer 32

Are you multiplying? -5 times -1?

Answer 33

I dunno

Answer 34

ok u said -5+i/2i= -5/2i + i/2i then that turns into -5/2i+1/2 then you loose me....

Answer 35

We apply the trick that I posted.
\[\large\rm \frac{1}{i}=-\frac{i}{1}\]Applying this trick tells us that\[\large\rm \frac{-5}{2i}=-\frac{-5i}{2}\]

Answer 36

ok yeah now i understand u take the i in the denominator and bring it up making it -5i and leaving the 2 alone ok

Answer 37

But notice now that it has two negatives, so let's just make that a positive.\[\large\rm \frac{5i}{2}\]

Answer 38

ok

Answer 39

\[\large\rm =\frac{5i}{2}+\frac{1}{2}\]This is an acceptable answer.
But let's apply a couple of steps to put it into "standard form".

Answer 40

ok

Answer 41

We would like to write the `real part first`.\[\large\rm =\frac{1}{2}+\frac{5i}{2}\]And then finally, let's pull the i off of the fraction so it's in the form a+bi.\[\large\rm =\frac{1}{2}+\frac{5}{2}i\]

Answer 42

One thing I'll say though...
The method I was first trying to show you,
multiplying top and bottom by something,
is much much better for these types of problems.

If you run into a problem such as\[\large\rm \frac{5+i}{2-i}\]You won't be able to do this same approach that we did here.

Answer 43

ok so my final answer would be 1/2+5/2i?

Answer 44

Yes.

Answer 45

ok thx lots! could you plse help me on one more question and thats it