do you have to remember all the formula for α and β (alpha and beta)

kevogy

there seems to be a new one ever year

Mongey

kevogy wrote: »

there seems to be a new one ever year

Just use the b rule to find the roots.

SligoBrewer

yeah, they can nearly all be deduced from α^2 and β^2 and α^3 and β^3.

You only have to remember that the sum of the roots is -b/a, and that the product is c/a.

From these two, (a+b) and (ab), every other one that they can ask can be derived with some simple ingenuity.

Swizz

kevogy wrote: »

there seems to be a new one ever year

No you just learn the a^3 + b^3 and the a^2 + b^2

if thats what your talkin bout otherwise

You need to know how to get the oots if theyre a and b as someone posted above me.

BLARG

Just know the one for alpha² + beta² and cubed.

Swizz

Yeah

Fringe

It only takes a minute at most to derive them anyway. There's no need to learn off something you don't need to. Just know how to get A and B.

Swizz

Fringe wrote: »

It only takes a minute at most to derive them anyway. There's no need to learn off something you don't need to. Just know how to get A and B.

Derive
alpha² + beta² and cubed




...You cant therefore learn them 2 off and ya'll be grand.

Of course you can derive them, it takes 30 seconds to derive (a^2 + b^2), and about 30 seconds more to get (a^3 + b^3). There isn't any point in trying to rote learn things which are, to be honest, simple to derive if you know how.

Swizz

-JammyDodger- wrote: »

Of course you can derive them, it takes 30 seconds to derive (a^2 + b^2), and about 30 seconds more to get (a^3 + b^3). There isn't any point in trying to rote learn things which are, to be honest, simple to derive if you know how.

Do it so.

Swizz

Swizz wrote: »

Do it so.

If you're told that the sum is 6, or something:

a + b = 6
(a + b)^2 = 36
a^2 + 2ab + b^2 = 36
(a^2 + b^2) = 36 - 2ab

You know what ab is (you're told), so, they're you go. That took me all of 30 seconds to type out.

For a^3 + b^3, it's just a matter of cubing instead of squaring what I've just done.

Swizz

-JammyDodger- wrote: »

If you're told that the sum is 6, or something:

a + b = 6
(a + b)^2 = 36
a^2 + 2ab + b^2 = 36
(a^2 + b^2) = 36 - 2ab

You know that ab is (you're told), so, they're you go. That took me all of 30 seconds to type out.

For a^3 + b^3, it's just a matter of cubing instead of squaring what I've just done.

NO. *PALMFACE*

challengemaster

Lol.

Know what alpha + beta is, know what alphabeta is. From that, you can easily find alpha² + beta²,alpha² - beta², and alpha³ + beta³, etc.

It's just a matter of knowing how to simplify and expand squares.

(X+Y)² = X² + Y² + 2xy
(X-Y)² = X² + Y² - 2xy
So on, so forth. Binomial theorem tbh.

Swizz

Swizz wrote: »

NO. *PALMFACE*

And what is wrong with what I've done?

What I've done is equal to:

a^2 + b^2 = (a+b)^2 - 2ab

It might write it differently in whatever book you've learned it from, but, I can tell you, they're equal.

challengemaster

Swizz wrote: »

NO. *PALMFACE*

What's wrong with that? Nothing, that's what. Stop stirring **** when you've got no idea what you're talking about.

Swizz

challengemaster wrote: »

What's wrong with that? Nothing, that's what. Stop stirring **** when you've got no idea what you're talking about.

Im not stirrng s**t, am trying to help the OP. I know what Im talking about. Ye dont.

You DO need to learn off a^3 + b^3 and a^2 + b^2. Ill find a Q after then Leaving and prove your dumbass wrong.

Swizz

Swizz wrote: »

Im not stirrng s**t, am trying to help the OP. I know what Im talking about. Ye dont.

You DO need to learn off a^3 + b^3 and a^2 + b^2. Ill find a Q after then Leaving and prove your dumbass wrong.

Well, if you think so.

But, you're wrong. Learn it off if you feel that's what works for you; but, it's possible to derive any possible combination of alpha and beta that they can ask.

PurpleFistMixer

You don't need to learn off a formula if you know how to derive it on the spot. Calm down with the aggro.

challengemaster

Swizz wrote: »

Im not stirrng s**t, am trying to help the OP. I know what Im talking about. Ye dont.

You DO need to learn off a^3 + b^3 and a^2 + b^2. Ill find a Q after then Leaving and prove your dumbass wrong.

PurpleFistMixer wrote: »

You don't need to learn off a formula if you know how to derive it on the spot. Calm down with the aggro.

Oh, snap!

Swizz

2008 Q2 b ii

You need the formula that Im talking about or you need to know(by learning off) the sum and diff. of 2 cubes.

Swizz

Swizz wrote: »

2008 Q2 b ii

You need the formula that Im talking about or you need to know(by learning off) the sum and diff. of 2 cubes.

You don't need to have learned either. If you understand how the sum and difference of two squares or cubes work, then, with a bit of thought you can derive whatever formulae that you may need.

Swizz

-JammyDodger- wrote: »

You don't need to have learned either. If you understand how the sum and difference of two squares or cubes work, then, with a bit of thought you can derive whatever formulae that you may need.

OMFG. Stop making a fool of yourself.

Fringe

You don't need to learn them off.

To get A^2 + B^2, you just find (A + ^2.

(A + ^2 = A^2 + 2AB + B^2
A^2 + B^2 = (A + ^2 - 2AB

Took me a few seconds to do it. Cube is a bit more tedious but it's just as easy.

Swizz

Swizz wrote: »

OMFG. Stop making a fool of yourself.

Where do you suppose the sum and difference formulae for cubes have came from? Out of thin air? No. They had to have been derived from somewhere (the result of this derivation was then place in your book, without showing the work). If you understand how they were derived, then, you can do the same thing. You don't have them.

Swizz

Look at the q the OP asked, do you really think hes gonna be able to derive that, no matter how simple that is, and theres a formula for a^3 + b^3 thats so simple and easy you cant get it wrong.

Swizz

Swizz wrote: »

Look at the q the OP asked, do you really think hes gonna be able to derive that, no matter how simple that is, and theres a formula for a^3 + b^3 thats so simple and easy you cant get it wrong.

2008, Question 2, part II: (without knowing any formulae)

(a+b)^3 = (a^2 + 2ab + b^2)(a+b)
= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3
a^3 + b^3 = (a+b)^3 - 2ab(a+b) -ab(a+b)

Therefore, 1/a^3 + 1/b^3 = (a^3 + b^3)/(ab)^3

a+b = 7
ab = 1

[7^3 - 2(1)(7) - 1(7)]/1^3
343 - 14 -7
322

Correct answer? Yes. Did I need pre-learned formulae? No.

Now, do you see that you were wrong?

It isn't too hard to derive.

Swizz

[quote=Swizz;60506142]Look at the q the OP asked, do you really think hes gonna be able to derive that, no matter how simple that is, and theres a formula for a^3 + b^3 thats so simple and easy you cant get it wrong.[/quote]
.

challengemaster

Swizz wrote: »

OMFG. Stop making a fool of yourself.

The only one doing that here is you. It's funny, about 5 people have now told you that you can derive them, and you still dont get it.

How do you think any cubed formula was got in the first place?



:pac:

papu

they arent that hard
if you try derive it and muck up your gonna drop marks

you dont have to remember it , but knowing both how to derive it and the equation will stop you drawing blanks on friday

I'm just proving the point that you were so adamant against: that you don't need to learn off any formulae. Calling other people a "fool" and a "dumbass" and insisting that they were wrong when, infact, they were right is quite ironic really.

Don't be so hostile to other people when they're showing you something which may prove benefitial to you.

Edit: Oh, and I did answer the OP's question. They were asking "do you have to remember all the formulae".