So yeah, my name is arx, and I've been a mathematical idiot for going on eight and three half years now...

I'm currently doing a maths c/w, and it involves quadratic equations, now I'm not going to post the exact problem because that's the idiots way out but I'd really appreciate it if someone was to give me a brief yet detailed tut on them, as well as an example.

Thanks for fuelling my desire to learn, better myself and more importantly - get more sleep tonight.

arx_

[This message has been edited by arx (edited 02-26-2006).]

Say you have an equation x^2+2x+1=0. You have to find two values that, when added to one another, make the coefficient of the middle term, and when multiplied, produce the last term. Then, you place these number inside brackets with x.

The solution in this case is (x+1)(x+1)=0. The zero, or solution, lies at x=-1.

It can get slightly more complicated, say in x^2-3x+2=0.

You would find two values that add to -3 and multiply to 2. The solution in this case is (x-2)(x-1)=0, and the x=2,1. The reason the signs change is because you make each bracket equal to zero and solve for x.

Many times it is more complicated than this, say in equations such as:

-2x^2+x+3=0

In this case, you may use a method that does not make sense algebraically, but works.

Multiply the last term by the coefficient on x^2, the first term. This gives us:

x^2+x-6=0

Then, you factor as you would normally, but put the coefficient back on both x's at the end:

(x-2)(x+3)

(-2x-2)(-2x+3)=0

Now, make both brackets equal to zero, and solve each bracket separately :

I.
-2x-2=0
-2x=2
x=1

II.< BR>-2x+3=0
-2x=-3
x=3/2< P>Therefore, the zeroes are at x=1 and x=3/2. The factored quadratic is (x-1)(x-3/2)=0.

If the quadratic is too tough for these tricks, use the formula.

In quadratic ax^2+bx+c=0, x=(b+sqrt(b^2-4ac))/2a, (b-sqrt(b^2-4ac))/2a

EDIT: Fucking dyslexic

[This message has been edited by McLaren (edited 02-27-2006).]

quadratic equations represent porabolas. zeros of a quadratic equation are the x-intercepts. There may be 0, 1, or 2 zeros. thats about it.

quote:

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Originally posted by McLaren:
Say you have an equation x^2+x+2=0. You have to find two values that, when added to one another, make the coefficient of the middle term, and when multiplied, produce the last term. Then, you place these number inside brackets with x.

The solution in this case is (x+1)(x+1)=0. The zero, or solution, lies at x=-1.

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I'm not a real math guy, lol, but it looks to me that (x+1)(x+1) would be x^2+2x+1.

If you wanted to check your work or if you just wanted to slack off. http://tinyurl.com/mgbsj

just plug the vaules in

quote:

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Originally posted by niggersexual:
I'm not a real math guy, lol, but it looks to me that (x+1)(x+1) would be x^2+2x+1.

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Silly me. I hate it when I do that.

x=-b � sqrt(b^2 - 4ac)/2a < quadratic formula

buy a TI-86 calculator (solves N order Polynomials)

Well yeah, perhaps a little difficult to understand but the idea is you factorise into something like (X + Y)(X + Z) = 0
As the "answer" is zero, one of the parenthesis must equal zero (so you get 0(XZ) = 0 or 0(XY) = 0).
So the answer will be -(Z) or -(Y), as -(Z) + Z = 0

(Or without the parenthesis for a variable..?)

I just remember a pretty cool funny thing. It's called slide-and-divide. You use it for when the squared term has a dumb coefficient.

Let's start out with (3x+5)(2x+7) just for example. This will give us the equation 6x^2+31x+35. Now let's try to solve it and get back to what we started with.

Multiply the constant term by the coefficient of the x^2 and just simplify the term to x^2.

Now we have x^2+31x+210

Now find the pair of factors of 210 that add up to 31. In this case, they're going to be 21 and 10.

Simplify to (x+10)(x+21).

Now are going to come the funny two steps.

Divide by the original coefficient of the squared term. Now we have (x+10/6)(x+21/6). Simplify all the way to get (x+5/3)(x+7/2). It has to be simplified all the way.

The second of the two funny parts and the most funny part comes next. Multiply the xs in each binomial by the denominatior to get (3x+5)(2x+7).

Now we can conclude that x={-5/3,-7/2}

I think this is a pretty good method. Using this trick, you'll become the king of your algebra class. Trust me.

[This message has been edited by niggersexual (edited 03-19-2006).]

Mr. Sexual:

You state that the x={-7,-5}, but they sould be x=-5/3 and x=-7/2 shouldn't they? I mean, I don't see why you do the denominator thing where you move it to the coefficient. Because if you just leave it, you just have to make it negative (because you set it equal to zero, and solver for x).

Not trying to argue, just wondering if my thinking in correct.

[This message has been edited by lanew (edited 03-19-2006).]

Also, another way to determin the x-intercepts, is to use "Decompostition", I am not sure if anyone is aware of this because they two methodes used for solving when there is a coefficient infron of the x squared are different.

That is basically where this method comes into play, when you have a coefficient infront of x squared.

y=3x^2-2x-8

Multiply the coefficient of x^2 by the single number without a variable. (3x-8=-24)

Take the answer (in our case, -24) and factor it:
-1,24
1,-24
-2,12
2,-12
-3,8
3,-8
-4,6
4,-6

From these, you take the pair of factors that add to give you the coefficent of x (the middle term). So for our example, it is -6 and 4, (-6+4=-2).

Now, take those numbers, and put them into the equation so you don't change it (bad wording, I know, but you'll see what I mean.)
3x^2 - 6x +4x - 8 (NOTE, I did not change the equation, it can be simplified to get back to the original equation).

Now, factor it, pull out common terms.

3x(x-2)+4(x-2)

Now, if you look at the equation, you can see there is a common term (x-2), factor that out, and you're done =).

(3x+4)(x-2)

There, it is now factored, from this, we can conclude that the roots are x=-4/3 and x=2.

YAY

quote:

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Originally posted by lanew:
Mr. Sexual:

You state that the x={-7,-5}, but they sould be x=-5/3 and x=-7/2 shouldn't they? I mean, I don't see why you do the denominator thing where you move it to the coefficient. Because if you just leave it, you just have to make it negative (because you set it equal to zero, and solver for x).

Not trying to argue, just wondering if my thinking in correct.

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Quite right. I suppose I wasn't thinking or something like that. Sorry if I caused any confusion.