what u mean by amplitue of a complex no?

- AravindG

what u mean by amplitue of a complex no?

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- AravindG

hi turing do u knw?

- AravindG

is it the modulus?

- TuringTest

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

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## More answers

- AravindG

ys

- AravindG

i thonk its modulus

- TuringTest

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

- AravindG

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

- TuringTest

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?

- AravindG

bt product should be -3??ryt?

- TuringTest

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

- TuringTest

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

- TuringTest

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

- AravindG

3x+2x?

- TuringTest

dang I had to reload, sorry...

- TuringTest

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?

- AravindG

ys

- TuringTest

let's try it...

- AravindG

k

- TuringTest

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...

- TuringTest

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

- AravindG

both r same

- TuringTest

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

- AravindG

hmm .. iam getting the taste of this method

- AravindG

we take 2x common then

- TuringTest

exactly

- AravindG

tastes good!!

- AravindG

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

- AravindG

turing?

- TuringTest

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

- AravindG

lol

- TuringTest

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

- AravindG

that was funny

- TuringTest

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

- TuringTest

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

- TuringTest

*
-3x-2x

- TuringTest

just take x common for the first pair

- AravindG

hey i got another way

- TuringTest

I know there are many
which are you referring to?

- AravindG

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

- AravindG

hws dat??

- TuringTest

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

- AravindG

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

- AravindG

-6x comes so +x

- AravindG

then -3 supplied hurray!!

- AravindG

hey i learnt to cook!!

- TuringTest

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

- AravindG

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

- AravindG

another qn i will post separately come there

- TuringTest

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

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