Question

I am sure my answer is correct although I was told it is wrong...anyone please confirm.
(4+3i)*(7-i)
My answer= -3i^2+17i+28

Answer 1

Just a minor mistake.. :(

Answer 2

remember what \(i^2\) is

Answer 3

oooh...so i am wrong :(

Answer 4

Sorry, not mistake but you have forgot to do further.. :)

Answer 5

\[i ^{2}= i \times i\]

Answer 6

We know that \(i^2 = i \cdot i\) but what is the value of \(i^2\) ??

Answer 7

i dont know..please explain

Answer 8

Oh God what I am doing today.. :) -1 there..

Answer 9

oook...so i= -1?

Answer 10

\[\large \color{green}{i^2 = -1}\]

Answer 11

since i is an imaginary number that is defined by \(i^2=-1\) look at your answer and you will see what else you need to do.

Answer 12

\(i = \sqrt{-1}\).. :)

Answer 13

is it like a rule ?

Answer 14

yes

Answer 15

ooooooh....i never knew that one

Answer 16

As long as you are studying complex numbers, then it must be a rule.. :)

Answer 17

Now i see why my answer was half way done!

Answer 18

as @waterineyes pointed out:
\[i=\sqrt{-1}\] thus
\[i^2=\sqrt{-1}^2 = -1\]

Answer 19

@Mokeira are you studying Complex Numbers as your topic??

Answer 20

@waterineyes I am!....teaching myself though that is why I was stuck

Answer 21

I mean to say that what topic you are doing?

Answer 22

Operations on Complex Numbers

Answer 23

Sub topic in calculus I think

Answer 24

Then you must know this :

Real numbers cannot define some numbers like : Negative Numbers under Square roots..

So, we have made another number set called Complex Numbers set. Complex Number itself starts from :

\[i = \sqrt{-1}\]

As sqrt{-1} cannot be defined by real numbers so, we have chosen our set as Complex and decided to take -ve numbers under sqrt in terms of iota ie i..

Answer 25

oooooooh wow! Thanks!!!!!! you are awesome. I have learnt something new

Answer 26

Like :

\[\large \color{green}{\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i}\]

Answer 27

ok wait.... why not -4i?

Answer 28

This way you can, any number, write in terms of i..

Answer 29

on the converse, squaring complex parts are done as such:
\[bi^2=-b^2\]
where b is a real number

Answer 30

Why?? \(\sqrt{-1}\) is \(i\).

Answer 31

now I get it....!

Answer 32

@waterineyes @agreene You guys ROCK!!! thank youuuuu I have learnt soo much

Answer 33

Keep on learning.. :)

Answer 34

np, that's why im here lol

Answer 35

so, getting back to your problem:
\[-3i^2+17i+28\]
\[3^2+17i+28\] and so on

Answer 36

Only \(3\) there..

Answer 37

Just a typing mistake nothing else.. :)

Answer 38

so can you guys also explain to me how The equation \[ax ^{2} + bx + c = 0\] can be rewritten (when a is not 0, after dividing by a) as
\[(x+\frac{ b }{ 2a }^{2})=\frac{ b ^{2}-4ac }{ 4a ^{2} }\]

Answer 39

Yes it can be written as like you said.. :)

Answer 40

i dont understand how you change from the equation to the fraction?

Answer 41

To scare you.. :P

Answer 42

lol... so its also like another rule?

Answer 43

No, it can be achieved by some mathematical steps.. :)

Answer 44

how...i have tried and cant figure it out

Answer 45

Have you heard of "Completion of Square Method" ??

Answer 46

it says divide by a

Answer 47

So what loss you are bearing by dividing by a?? Just do it friend.. :P

Answer 48

No I havnt...let me read about it then I try again

Answer 49

Okay jokes apart... :) let us do it now :)

Answer 50

Actually I know the "Completion of square method"

Answer 51

Can you divide by a first?? As you are saying.. :)

Answer 52

ok..let me divide by a

Answer 53

If you know then it will solve your problem.. :)

Answer 54

Oh my God @UnkleRhaukus uncle rocks is here only.. :) How are you Uncle Rocks??

Answer 55

hi !

Answer 56

\[\frac{ a ^{2+bx+c} }{ a }=a+\frac{ bx }{ a }+\frac{ c }{ a }\]
Am I right?

Answer 57

Hi, Really I am very happy to see you here.. :) Long time..

Answer 58

You should check it once again.. :)

Answer 59

You have to divide Left Hand Side as well as Right Hand side by a..

Answer 60

the equation to be divided is \[a ^{2}+bx+c\]
there is no right and left hand side

Answer 61

sorry..not an equation

Answer 62

If no right hand side, then why are you saying it as equation??

Answer 63

not an equation..i corrected myself..

Answer 64

Okay, I am late.. :P

Answer 65

oh wait!...its =0

Answer 66

You are having an equation, Okay?/

\(ax^2 + bx + c = 0\)

Answer 67

Are you alright @Mokeira ?? Something wrong??

Answer 68

so even after you divide by a it will still be =0

Answer 69

Yes, but check for Left hand side once again.. :)

Answer 70

Can you clearly write down what have you got after dividing by a, write if fully with LHS and RHS like an equation.. :)

Answer 71

ok...sure. Dont get tired lol :)

Answer 72

He he he he.. Who is getting tired?? :)

Answer 73

\[a ^{2}+bx+c=0\]
\[\frac{ a ^{2}+bx+c=0 }{ a }\]
\[a+\frac{ bx }{ a }+\frac{ c }{ a }=0\]
Is this right

Answer 74

I hope you are not kidding this time..

Are you hungry??

Answer 75

Where is \(x^2\) ??

Answer 76

we are dividing by a right?
How and where does x^2 come from?

Answer 77

\[\frac{ax^2 + bx + c}{a} = 0 \implies x^2 + \frac{b}{a}x + \frac{c}{a} = 0\]

Answer 78

Dear, you should check your equation first.. :) See it with your eyeballs for 5 minutes.. :P

Answer 79

lol! I always omit!!!! i guess im always in a hurry. my bad! I get it now! Clumsy me

Answer 80

lol...thanks! thank u for your time too

Answer 81

can you go ahead with this now? Just use Completing the square method you will reach there where you want.. :)

Answer 82

It is okay.. :)

You are welcome dear.. :)

Answer 83

I have to go, if you have any doubt, then @UnkleRhaukus will help you here from nowonwards.. Keep doing more and more.. :)

Answer 84

ok! thanks bye

Answer 85

See you later @UnkleRhaukus :)

Answer 86

see you next time @waterineyes !