Question

Hey I need help with some questions they should be pretty easy. Any and all help appreciated.

Answer 1

\[\sqrt{-32}-\sqrt{-18}\]
Find the exact value.

Answer 2

Can you guys help me?

Answer 3

@trueseminole @jcpd910

Answer 4

Hmmm

Answer 5

Okay, I think the exact value would be the square root of 32 i.
Like this:
5.65i

Answer 6

Why the decimals? :c

Answer 7

Because the square root of a negative number is imaginary. I've given up trying to understand why, it just is.

Answer 8

Well he wants the exact value, so might as well try and get close.

Answer 9

Alright
But my prof says i need to show work :/

Answer 10

\[\sqrt{-32} = 4i \sqrt2 \]This is exact.

Answer 11

:D thanks, how about the rest of the problem?

Answer 12

@jcpd910 : Ring a bell?

Answer 13

Yes, but can you explain it to me?

Answer 14

Sure.

Answer 15

\[\sqrt{-32} = i \sqrt{32} =i \sqrt{16*2}=i \sqrt{4*4*2}=4i \sqrt{2}\]

Answer 16

Oh yeah, because sqrt -32 is sqrt -1 * 32!

Answer 17

\[\sqrt{-32}-\sqrt{-18}=4 i \sqrt{2}-3 i \sqrt{2}=\sqrt{2} (4 i-3 i)=i \sqrt{2} \]

Answer 18

Yes, sir.

Answer 19

But one thing, how does i sqrt 4*4*2 become 4i sqrt 2

Answer 20

alright, so if i were to do the same thing for the rest of the equation

Answer 21

Whenever there's a pair of numbers inside the root, one of them comes out.
Like \[\sqrt4 = \sqrt{2*2} = 2\]or
\[\sqrt18 = \sqrt{3*3*2} = 3 \sqrt2 \]

Answer 22

You beat me to it

Answer 23

Okay, so can you throw me a practice problem?

Answer 24

So would the
\[\sqrt{-18}\]
equal \[3i \sqrt{2}\] ?

Answer 25

https://www.khanacademy.org/math/pre-algebra/exponents-radicals/radical-radicals/e/multiplying_radicals
http://hotmath.com/help/gt/genericalg1/section_8_1.html

Answer 26

\(\bf \sqrt{-32}-\sqrt{-18}
\\ \quad \\
{\color{brown}{ 32\to 2\cdot 2\cdot 2\cdot 2\cdot 2\to 2^4\cdot 2\to (2^2)^2\cdot 2}}
\\ \quad \\
{\color{brown}{ 18\to 2\cdot 3\cdot 3\to 2\cdot 3^2}}\qquad thus
\\ \quad \\
\sqrt{-32}-\sqrt{-18}\implies \sqrt{-1\cdot 32}-\sqrt{-1\cdot 18}\implies \sqrt{-1}\cdot \sqrt{32}-\sqrt{-1}\cdot \sqrt{18}
\\ \quad \\
i\ \sqrt{(2^2)^2\cdot 2}-i\ \sqrt{2\cdot 3^2}\)

Answer 27

@saifoo.khan hey just one more problem

Answer 28

Shoot.

Answer 29

\[\frac{ 3 + 4i }{ 4- 2i}\]
Simplify and Rationalize.

Answer 30

@jcpd910 wanna try?

Answer 31

\[\frac{3+4i}{4-2i} \times \frac{4+2i}{4+2i} \]

Answer 32

Can you divide complex numbers? Like: 4i/-2i = -2i?

Answer 33

Okay I'll try.

Answer 34

And nevermind saifoo o_o
But alright, once I have what you just did, how do I proceed?

Answer 35

\[\frac{ (3+4i)(4+2i) }{ (4-2i)(4+2i) } \times \frac{ (3+4i)(4+2i) }{ (4-2i)(4+2i) }\]
Right so far?

Answer 36

Wait

Answer 37

\[\frac{ (3+4i)(4+2i) }{ (4-2i)(4+2i) } \times \frac{ (3+4i)(4-2i) }{ (4-2i)(4+2i) }\]

Answer 38

There, correct so far?

Answer 39

How did you...
I don't understand

Answer 40

I multiplied both things by the other denominator in order to get common denominatros.

Answer 41

Hmm

Answer 42

But if you do that wouldn't you only get the first half of what you got?

Answer 43

O_O

Answer 44

Lol saifoo whats going on

Answer 45

lol yeah derp

Answer 46

I DONT EVEN NKOW!

Answer 47

halp meeeee *slowly dies*

Answer 48

@saifoo.khan Help us wise one.

Answer 49

wait
isn't is just:
\[\frac{3+4i}{4-2i} \times1\]

Answer 50

In which case isn't it just:
\[\frac{3+4i}{4-2i} \]

Answer 51

You'd end up with the same exact equation you started with though....Wouldnt you?

Answer 52

Lol what?!

Answer 53

Lol. Okay so:
\[\frac{ 3 + 4i }{ 4- 2i}\]Rationalize.
\[\frac{3+4i}{4-2i} \times \frac{4+2i}{4+2i}\]
Now multiply.
\[\frac{(3+4i)(4+2i)}{(4-2i)(4+2i)}\]

Answer 54

So you do nothing?!

Answer 55

Ahhhh

Answer 56

Oh I thought we started off different

Answer 57

Now just use FOIL.

Answer 58

Lol jcpd you did some crazy **** I was wondering where you got so many numbers

Answer 59

lerl brb

Answer 60

And alright saifoo I'll try it.

Answer 61

Alright does it look something like this: