quadratic eqations

can any one help me with factorizing quadratic equations.

Quote:

---

B2 - 5B = 6

---

i dont understand how to work them out when they look like that.

any help would be great :)

thanks

bring them to one side, so you have:
B²-5B-6=0

you are trying to find two numbers that add up to 5 and multiply to get -6.

obviously -6 and 1 works (should be divisible by 6).

so it would factor into:

(B-6)(B+1)=0

thanks :)

for any quadratic equation ax²+bx+c=0
if b²-4ac>0
then you can split into these factors:
x+(b+sqr(b²-4ac))/2a
x+(b-sqr(b²-4ac))/2a

I think you got the formula wrong. It is:

x= (-b+sqrt(b^2-4ac))/2a
x= (-b-sqrt(b^2-4ac))/2a

And factoring is always much better. The quadratic formula is a last resort.;)

Bugz,

You said "bring them to one side"!!! Have you forgotten BS??? How unorthodox!

Frankly Kalkewl8ter, i dont think he cares.

WOW what a good job ms. romiens did!! LOL!

MS ZIEBA SUX SUX SUX SUX SUX

Quote:

---

Originally posted by SilverSprite
I think you got the formula wrong. It is:

x= (-b+sqrt(b^2-4ac))/2a
x= (-b-sqrt(b^2-4ac))/2a

And factoring is always much better. The quadratic formula is a last resort.;)

---

I said those were the factors {X-y|yÎ {x|ax²+bx+c=0} }

Okok I'm sorry.:(

MRS. ROMIENS IS EVIL!!!
She goes against Mr. White's ways, and you can't say Mr. White sucks, bugz...

why what does she do?

besides, so does mr. caldwell... he says "cross-multiplying"! does that mean mr. caldwell is evil too? of course not!

only mr.braithwaite doesn't do that, but he was mr. white's student!

I found a quick way of factorising quadratics with a coeficient of x that is greater than one. For example:

6x²+19X+10

[list=1][*] Find two numbers which multiply together to give the product of the constant (10) and the coeficient of x² (6) and add together to give the coeficient of x.

In this case the product of the constant and the coeficient of x² is:
60

And two numbers which add together to give the coeficient of x and multiply together to give 60 are:

4 and 15
[*] Rewirte the equation with the x coefitients split :
6x²+4x+15x+10
[*]Now take the commmon factors out and rewrite as:

3x(2x+5) +2(2x+5)
[*] The two factors can now be extracted visually. The answer is correct if both the factors in brackets match, in this case (2x+5) which is also a factor. The other factor can now be extracted and is (3x+2) - its split among the 2 lots of (2x+5)'s.[/list=1]

Solved completely this gives x a value of -5/2 and -2/3.

I'm yet to find a quick way of solving quadratics with an x² coeficient that is less than 0 e.g. -5 of something. So if anyone know of a way I'd be interested to know.

Factor out the negative sign, and you'll have a positive coefficient.
Doesn't anyone factor by math anymore?

I do occasionally :D

Occasionally? Mr White would be disappointed...you should do it all the time, bugz...
And so what if Mr Caldwell says cross-multiply? Mrs Romiens doesn't do BS, doesn't mind SOHCAHTOA, uses rise/run for slope, and likes y=mx+b!!!

I prefer to remember SOHCAHTOA as:

Sex On Holiday Can Affect Health Till Old Age:D

Mr. White doesn't seem to mind Mrs Romiens so why would I mind? and why do you take it so personally? to enforce some ego of yours that says if i got bad on tests, it must be teachers fault cause i am faultless?

Actually, bugz, I never even thought of that; I was just defending math. :) Mr White also doesn't seem to know how Mrs Romiens teaches...
(And I'll remember that one, visualAd.)

Thats such a noble cause! if you really honestly want to defend math, instead of talking about her behind her back, next time she doesn't uses BS, uses SOCAHTOA, confront her! Remember this, if you don't, that means you are just lying to yourself [and to us], and that you aren't really defending math, but defending something else.

We don't treat teachers etc like blood relatives over here so I'm getting lost in what's going in...

thats where you are wrong A$$Bandit! We are in fact all related!

We can all be traced back into the first human that evolved from primapes, according to the Charles Darwin's Theory of Evolution.

Do you know how to calculate the number of nodes at the nth level of a binary tree, A$$Bandit?

Code:

---

      0
      /\
      0  0
    /\  /\
    0 0 0 0

---

at the nth level (the root or top level being 0), it can be verified that there are 2^n nodes.

now imagine you are the root. you have two parents, they will be the nodes on the first level. your parents have two parents each, they will be nodes on the second level, and so on. That means n generations ago, you would have 2^n grandparents. assume each generation is 30 years, then 1500 years ago, or 50 generations ago, you would have 2^50 grandparents, or 1,125,899,906,842,624 (more than one quadrillion)!

Kalkewl8ter exagerates things way too much. You should let people be lazy sometimes, its for their own good. Too much work isnt good you know.

You cannot call them all grandparents as with each generation you need to add thee word Great. So the grandparents of three generations would be:

Great Great Grandparents

but the title "great great grandparents" still contains the word grandparents, with a sort of prefix "great great" therefore i can call them grandparents, or even parents if i wish.

Blood relatives? I never thought of it that way, A$$Bandit. And of course bugz sees math in all, but screws up the English part...grandparents=parents...
Therefore (divide BS by parents) grand=1!
And bugz, I do confront her about it...in fact I criticize her all through class. Sprite can be witness to that if he wants...but I doubt that he will.

Yeah right, all i hear is you mutter under your breath. (which could be criticizing). You never speak out.:P

HAHAHA thats what i thought :D

Of course I speak out! Mostly to Yang, but I yell out "difference of squares!" every other day, and I'm ever-antagonist against SOHCAHTOA and rise/run slope. She wouldn't explain what a canyon was in class! :mad: