Question

Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)

Answer 1

I know that the sqrt of -49 factors into i49.

Answer 2

i sqrt 49 is what I meant

Answer 3

\(\bf \cfrac{\sqrt{-49}}{(7 - 2i) - (4 + 9i)}\\ \quad \\
\textit{notice that }\quad \sqrt{-49}\implies \sqrt{49}\times \sqrt{-1}\\ \quad \\
\textit{let us use the }conjugate\textit{ of the denominator } (7 - 2i) + (4 + 9i)\\
\cfrac{\sqrt{-49}}{(7 - 2i) - (4 + 9i)} \times \cfrac{(7 - 2i) + (4 + 9i)}{(7 - 2i) + (4 + 9i)}\\ \quad \\
\cfrac{\sqrt{-49}[(7 - 2i) + (4 + 9i)]}{[(7 - 2i) - (4 + 9i)]\quad [(7 - 2i) + (4 + 9i)]} \\ \quad \\
\textit{keep in mind that } (a-b)(a+b) = (a^2-b^2)\\ \quad \\
\cfrac{\sqrt{-49}[(7 - 2i) + (4 + 9i)]}{[(7 - 2i) - (4 + 9i)]\quad [(7 - 2i) + (4 + 9i)]}\\\quad \\
\implies \cfrac{\sqrt{-49}[(7 - 2i) + (4 + 9i)]}{(7 - 2i)^2 - (4 + 9i)^2}\)

Answer 4

It kind of makes sense, but I thought you had to make sqrt -49 into a positive

Answer 5

you'd eventually, yes

Answer 6

I'm still confused though :l

Answer 7

oh okay

Answer 8

ok.... where? you still need to simplify it some, expand the bottom and top

Answer 9

.... ohhh... smokes... wait a second

Answer 10

I don't understand the 3rd step. I just don't remember all of that in my lessons

Answer 11

Okay (:

Answer 12

... all that was so superfluos.... shoot the denominator is easier simplifyable

Answer 13

lemme redo all that hehe

Answer 14

lol Okayy

Answer 15

\(\bf \cfrac{\sqrt{-49}}{(7 - 2i) - (4 + 9i)}\\ \quad \\
\textit{notice that }\quad \sqrt{-49}\implies \sqrt{49}i\implies \sqrt{7^2}i\implies 7i\\ \quad \\
\cfrac{\sqrt{-49}}{7 - 2i - 4 - 9i}\implies \cfrac{7i}{3 - 7i}\\ \quad \\
\textit{now let's use the conjugate of the denominator of }\quad 3+7i\\ \quad \\
\cfrac{7i}{3 - 7i}\times \cfrac{3+7i}{3+7i} \implies \cfrac{7i(3+7i)}{(3-7i)(3+7i)}\\ \quad \\
\textit{now recall that }\quad (a-b)(a+b) = (a^2-b^2)\\ \quad \\
\cfrac{7i(3+7i)}{(3-7i)(3+7i)}\implies \cfrac{7i(3+7i)}{(3)^2-(7i)^2}\implies \cfrac{21i+49i^2}{9-7^2i^2}\)

Answer 16

acck... hold.... I have a slight error there :(

-2i -9i = -11i
lemme fix that quick

Answer 17

its okay (: I forgot about seeing if -49 had any factors

Answer 18

\(\bf \cfrac{\sqrt{-49}}{(7 - 2i) - (4 + 9i)}\\ \quad \\
\textit{notice that }\quad \sqrt{-49}\implies \sqrt{49}i\implies \sqrt{7^2}i\implies 7i\\ \quad \\
\cfrac{\sqrt{-49}}{7 - 2i - 4 - 9i}\implies \cfrac{7i}{3 - 11i}\\ \quad \\
\textit{now let's use the conjugate of the denominator of }\quad 3+11i\\ \quad \\
\cfrac{7i}{3 - 11i}\times \cfrac{3+11i}{3+11i} \implies \cfrac{7i(3+11i)}{(3-11i)(3+11i)}\\ \quad \\
\textit{now recall that }\quad (a-b)(a+b) = (a^2-b^2)\\ \quad \\
\cfrac{7i(3+11i)}{(3-11i)(3+11i)}\implies \cfrac{7i(3+11i)}{(3)^2-(11i)^2}\implies \cfrac{21i+77i^2}{9-11^2i^2}\)

Answer 19

That's the final answer?

Answer 20

and when expanding the denominator, don't forget that \(\bf i^2\implies \sqrt{-1}\times \sqrt{-1}\implies \sqrt{(-1)^2}\implies -1\)

Answer 21

\(\bf \cfrac{21i+77i^2}{9-11^2i^2}\implies \cfrac{21i+77(-1)}{9-11^2(-1)}\)

Answer 22

Okay, but that's not any of the answer choices

Answer 23

well, what would you get for \(\bf \cfrac{21i+77(-1)}{9-11^2(-1)}\)

Answer 24

what do you mean?

Answer 25

if you expand both top and bottom

Answer 26

I don't know ):

Answer 27

what's \(\bf 77 \times -1\quad ?\)

what's \(\bf -11^2 \times -1\quad ?\)

Answer 28

Just making sure, is x suppose to be a variable or a multiplication sign? lol

Answer 29

multiplication :)
or maybe better \(\bf 77 \cdot -1\\
-11^2 \cdot -1 \quad ?\)

Answer 30

That's better lol -77 and 11^2?

Answer 31

right and 11^2 = 121
so we can say that \(\bf \cfrac{21i+77i^2}{9-11^2i^2}\implies \cfrac{21i+77(-1)}{9-11^2(-1)}
\implies \cfrac{21i-77}{9+121} \implies \cfrac{-77+21i}{130} \)

Answer 32

Oh, okay! That makes sense!

Answer 33

Can you help me out with another one? It's simple. I just forgot the first step lol

Answer 34

ok

Answer 35

Simplify 4 sqrt 192

Answer 36

I have to factor 192 right?

Answer 37

yes

Answer 38

okay, that's all I needed to know. Thanks so much!

Answer 39

\(\bf 192 \implies 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\implies 2^2\cdot 2^2\cdot 2^2\cdot 3\\\implies (2^2)^3\cdot 3\implies (2^3)^2\cdot 3\\ \quad \\
4\sqrt{192} \implies 4\sqrt{(2^3)^2\cdot 3}\)