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hi turing do u knw?
is it the modulus?
are you sure you don't mean 'magnitude'?
i thonk its modulus
yes modulus=magnitude though I can find nothing that says amplitude=modulus so I can't say for sure
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
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?
bt product should be -3??ryt?
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
so just read my post again and try to answer my question, please :)
look at the options on how we can rewrite this expression....
dang I had to reload, sorry...
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?
let's try it...
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...
try breaking the middle term up as\[-2x-3x\]
both r same
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
hmm .. iam getting the taste of this method
we take 2x common then
so we split middle term so that we get 'a'x in one of the parts
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
honestly I learned this method here! lol I always did it by eye before, so I sort of haven't practiced it enough
that was funny
so hell, what am I missing? I have to work it backwards....
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
just take x common for the first pair
hey i got another way
I know there are many which are you referring to?
it's correct and how did you come to that?
simple u see i understud the othr one wont work so i took 2x^2 and -3x and supplied neccessary terms
-6x comes so +x
then -3 supplied hurray!!
hey i learnt to cook!!
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
ya not exactly trial and error but i think it will be intutiuve after doing more such problems.anyway i learnt the method thanks
another qn i will post separately come there
Like I said the first time, no substitute for practice to develop intuition