Question

what u mean by amplitue of a complex no?

Answer 1

hi turing do u knw?

Answer 2

is it the modulus?

Answer 3

are you sure you don't mean 'magnitude'?

Answer 4

ys

Answer 5

i thonk its modulus

Answer 6

yes
modulus=magnitude
though I can find nothing that says amplitude=modulus so I can't say for sure

Answer 7

try this:
tell me general steps to be followed while factorizing quadratic eqns with values of a other than 1 like 2x^2-5x-3

Answer 8

it's all about the middle term
you want to break it up so you can factor out the GFC of each pair of terms\[ax^2+bx+c=ax^2+nx+mx+c\]where\[n+m=b\]5x can be written as
4x+x
3x+2x
2x+3x
x+4x
which of these choices above will allow you two factor something out of each pair of terms?

Answer 9

bt product should be -3??ryt?

Answer 10

yes, but I'm trying to teach you a technique called "factoring by grouping" that works for quadratics where the coefficient if the x^2 term is not 1

Answer 11

so just read my post again and try to answer my question, please :)

Answer 12

look at the options on how we can rewrite this expression....

Answer 13

3x+2x?

Answer 14

dang I had to reload, sorry...

Answer 15

anyway, I dropped th negative, so the options are
-5x=
-x-4x
-2x-3x
-3x-2x
-4x-x
so your choice is that we should break it up as
-3x-2x
you say?

Answer 16

ys

Answer 17

let's try it...

Answer 18

k

Answer 19

plugging that in gives\[2x^2-3x-2x-3\]now look at each \(pair\) of terms...\[2x^2-3x\]leaves us only able to factor an x out\[-2x-3\]leaves us only able to factor out a -1
we can do better than that...

Answer 20

try breaking the middle term up as\[-2x-3x\]

Answer 21

both r same

Answer 22

technically yes, but writing it in this particular way will reveal how to factor this equation
that is the nature of the "factoring by grouping" technique

Answer 23

hmm .. iam getting the taste of this method

Answer 24

we take 2x common then

Answer 25

exactly

Answer 26

tastes good!!

Answer 27

so we split middle term so that we get 'a'x in one of the parts

Answer 28

turing?

Answer 29

yeah I'm sorry...
the idea I'm talking about is correct, but I'm pretty sure you are having the same problem that I am with the second pair of terms

Answer 30

lol

Answer 31

honestly I learned this method here! lol
I always did it by eye before, so I sort of haven't practiced it enough

Answer 32

that was funny

Answer 33

so hell, what am I missing?
I have to work it backwards....

Answer 34

you know what, you had it right the first time
I was misrepresenting the goal, sorry again
use
3x+2x
and take -1 common from the second pair

Answer 35

*
-3x-2x

Answer 36

just take x common for the first pair

Answer 37

hey i got another way

Answer 38

I know there are many
which are you referring to?

Answer 39

2x(x-3)+(x-3)

Answer 40

hws dat??

Answer 41

it's correct
and how did you come to that?

Answer 42

simple u see i understud the othr one wont work so i took 2x^2 and -3x and supplied neccessary terms

Answer 43

-6x comes so +x

Answer 44

then -3 supplied hurray!!

Answer 45

hey i learnt to cook!!

Answer 46

yeah, I think you're sort of describing the way I see it
I do a little trial and error and you figure out which gives you the right thing to factor out

Answer 47

ya not exactly trial and error but i think it will be intutiuve after doing more such problems.anyway i learnt the method thanks

Answer 48

another qn i will post separately come there

Answer 49

Like I said the first time, no substitute for practice to develop intuition