Question

Iam completely lost on doing the problem can someone plz explain to me how to do it.
simplify the given expression
sqrt-64/(7-6i)-(2-2i)

Answer 1

Sure :)
\[\Large \frac{\sqrt{-64}}{(7-6\color{red}i)-(2-2\color{red}i)}\]

Answer 2

So, what is the square root of 64?

Answer 3

i know i have to distribute

Answer 4

distribute? Like in the denominator?
Well, go ahead, then :)

Answer 5

but he sqrt of 64 is 8

Answer 6

That's right... but it's -64
What's the square root of -64?

Answer 7

-8?

Answer 8

No... because -8 times -8 is still 64 :)
We're talking about complex numbers anyway, the square root of a negative number is
i times the square root of the number.
for instance
\[\Large \sqrt{-25}= 5\color{red}i\]

Answer 9

That said, what's the square root of -64?

Answer 10

\[\sqrt{-64}=8i\]

Answer 11

You catch on quickly :)
So let's replace the numerator...
\[\Large \frac{8\color{red}i}{(7-6\color{red}i)-(2-2\color{red}i)}\]
Now simplify the denominator... you can do that by combining like terms, much like in algebra :)

Answer 12

8i/14-14i-12i+12i^2

Answer 13

OH... there is no multiplication involved (yet)
It's just adding and subtracting

Answer 14

Distribute the negative sign first, to make things easier.
\[\Large \frac{8\color{red}i}{(7-6\color{red}i)\color{green}-(2-2\color{red}i)}\]

Answer 15

-2+2i

Answer 16

Great :)
so it now becomes
\[\Large \frac{8\color{red}i}{7-6\color{red}i-2+2\color{red}i}\]
combine like terms in the denominator, you get?

Answer 17

yes
5-4i

Answer 18

Brilliant :)
\[\Large \frac{8\color{red}i}{5-4\color{red}i}\]
Now, much like radicals, people don't like imaginary units in the denominator...

Answer 19

so i multipliy by i correct?

Answer 20

A way to get rid of that imaginary unit is multiply both the numerator and denominator by the CONJUGATE of the denominator.

You know what a conjugate is? :)

Answer 21

yes means you use the opposite sign

Answer 22

(5+4i)

Answer 23

Very nice :)

Answer 24

Multiply both the numerator and denominator by that conjuate
\[\Large \frac{8\color{red}i\cdot(5+4\color{red}i)}{(5-4\color{red}i)(5+4\color{red}i)}\]

Answer 25

conjugate*

Answer 26

40i+32i^2/25i

Answer 27

Whoa now... work on that denominator again...
\[\Large (a-b)(a+b)=a^2-b^2\]

Answer 28

ok
wil it be -3

Answer 29

Nope :)
Look at it, it's a sum and difference...
Use this form \[\Large (a-b)(a+b)=a^2-b^2\]
where a = 5
b = 4i

Answer 30

(5-4i)(5+4i)=25-16i^2

Answer 31

Good, now just remember that
\[\Large \color{red}i^2 = -1\]

Answer 32

ok
32+40i/-1

Answer 33

What? LOL don't be so hasty.
\[\Large 25 - 16\color{red}i^2 = 25 - 16(\color{red}{-1})\]
simplify

Answer 34

i am confused:/

Answer 35

will it be25-16i=25+16

Answer 36

Precisely :)
Remember, the whole reason we multiplied both the numerator and denominator by the conjugate...

Answer 37

WAS TO GET RID OF THE \(\large \color{red}i\) in the denominator !!!

Answer 38

oh ok

Answer 39

So you understand it now :)
At the numerator, what did you get?

Answer 40

i do,but the only answer choice i have that is close to this is
32+40i/41

Answer 41

is that right

Answer 42

Check the numerator...

Answer 43

\[\Large 8\color{red}i(5+4\color{red}i)\]
distribute?

Answer 44

40i+32i^2

Answer 45

wich gives me -31

Answer 46

Interesting.
And it's -32 btw :)

Answer 47

What are the answer choices?

Answer 48

on sec

Answer 49

ok.
a.-64+40i/-3
b.34+40i/41
c.64+40i/-3
d.-32+40i/41

Answer 50

smh... there's a -32 in choice (d)
:P

Answer 51

thanks for your help i am not very good in math

Answer 52

You actually are.
Just need to avoid that spark of carelessness every now and then :)

Answer 53

lol thanks can you check two more for me to see if i did them correctly? if you have time.

Answer 54

Hit me

Answer 55

thank you so much

Answer 56

sqrt-6(2+sqrt8)
a.2sqrt6+4sqrt3
b.-2sqrt6+4sqrt3
c.-4sqrt3+2isqrt6
d.4sqrt3+2isqrt6
i chose d

Answer 57

Correct :)

Answer 58

ok last one

Answer 59

sqrt-10*sqrt-8
a.-4sqrt5
b.4sqrt5
c.4isqrt5
d.-4isqrt5
i chose c

Answer 60

wrong :)
remember that multiplying through radicals is only valid if the radicands are positive, which they aren't :)

Answer 61

so the 4 has to be a negative correct

Answer 62

Yup :)

Answer 63

ok thanks so much i feel beteoing these now
:)

Answer 64

wait.
what's your answer?

Answer 65

sorry d is my answer

Answer 66

still no :)
Look, I'm tired, so I'll run it through you in as much detail as possible.
Try to study the steps thoroughly, okay? :)

Answer 67

ok thank u

Answer 68

\[\Large \sqrt{-10}\sqrt{-8}\]

Answer 69

So, we take out the i's...
\[\Large (\color{red}{i}\sqrt{10})(\color{red}i\sqrt{8})\]\[\Large \color{red}i^2\sqrt{10}\sqrt8\]
\[\Large \color{red}i^2\sqrt{40}\]

Answer 70

Again, let's not forget that \[\Large \color{red}i^2 = -1\]

Answer 71

Whoops, typo. Sorry..
\[\Large \color{red}i^2\sqrt{\color{blue}{80}}\]

Answer 72

Now... \(\large \color{red}i^2 = -1\)
\[\Large (-1)\sqrt{16\cdot 5}\]
\[\Large (-1)4\sqrt5\]
\[\Huge\color{blue}{-4\sqrt5}\]

Answer 73

oh ok i see now thanks for ur help

Answer 74

Practice okay?
Promise... ;)

Answer 75

i promise

Answer 76

That's all I need to know :)
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Terence out