Question

radical

Answer 1

@DanJS

Answer 2

k

Answer 3

\[(-3\sqrt{-5})(-3\sqrt{-5})\]

Answer 4

or
\[(-3\sqrt{5})^2\]

Answer 5

yeah, just think of it as multiplication of all 4 of the terms

Answer 6

-3*-3*root5*root5

Answer 7

\[(-3*-3)*\sqrt{5*5} = 9*\sqrt{25} = 9*5 = 45\]

Answer 8

\[9\sqrt{25}\]

Answer 9

yep

Answer 10

wait..but you got 45??

Answer 11

square root of 25 is 5

Answer 12

o wait you have -5 in there,

Answer 13

there will be some of those i's involved because of the negative under the radical

Answer 14

\[(-3*-3)\sqrt{-1*5}*\sqrt{-1*5}\]

Answer 15

\[9*\sqrt{-1}\sqrt{-1}\sqrt{25}\]

Answer 16

remember, i = root(-1)

Answer 17

\[9 * i*i * 5\]

Answer 18

45i^2

and i^2 = -1

45*-1

-45

Answer 19

Okay, I see it now. So, there is not an i in the answer correct? just -45?

Answer 20

right, only an i if the i has an odd power

i^1=i
i^2=-1
i^3=-i
i^4=1

Answer 21

okay

Answer 22

switching gears for a sec....what is a conjugate?

Answer 23

conjugate pair i think , remember those denominators that had

\[a +\sqrt{b}\]

Answer 24

then you multiply that by the conjugate

\[a - \sqrt{b}\]

Answer 25

http://www.mathsisfun.com/algebra/conjugate.html

Answer 26

okay. I thought it was where you switch the signs but I wasn't sure.

Answer 27

yeah, it is any binomial ( a thing that has 2 terms), with the sign switched

Answer 28

binomial a + root(b)

conjugate a - root(b)

Answer 29

thanks.

Answer 30

I only have a few more questions that I am stumped on if you could help....I normally wouldn't ask you to help with a lot of questions but I want to be able to be prepared for my test. Afterwords...I won't be bothering you anymore. :)

Answer 31

no prob, start a new thread so i can get more medals. lol

Answer 32

haha :)