Talk:Quadratic equation/Archive 2

This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Archive 1

Archive 2

Archive 3

Archive 4

Archive 5

Descriminate and factorisation

Added effect of descriminate on factorisation of quadratics. User:Wolfmankurd 17:19, 16 May 2006 (UTC)

...

In English please? ForestAngel 10:47, 16 August 2007 (UTC)

Factorisable

Am I right in thinking the (%) of quadratics that are unfactorisable limiuts to 100%? Based on the fact that the descriminate must be a positive square, and the square numbers have bigger and bigger gaps inbetween. So if you have quadratics where a, b and c are between -n and n then as n limits to infinity the (%) of factoriseable equations limits to 0? Sorry it sounds so confusing I cant explain it properly. Partly why I havent included it my self. Wolfmankurd 21:41, 7 June 2006 (UTC)

It's certainly true that there are more quadratics that factor over the integers or rationals than that factor over the reals or complexes. The former are countable, the latter are uncountable. Certainly then the former make up a set of measure zero in the space of quadratics, so yes, it would be fair to say 100% of them, in the sense that if you threw a dart at them, it would land on an unfactorable quadratic with probability 1. -lethe ^{talk +} 23:08, 7 June 2006 (UTC)

toper degrees

One of the issues I have with math text books is that at first I am thrilled to be learning this knowledge, and I am happy to be connected with truth, harmony, and the secrets of the universe, but then there's a quick switcheroo, and I am lost and confused, and I feel as if I am on the wrong side of an academic three card monte game. Now I see it, now I don't, and now I'm the chump.

For example, the into paragraphs on the quadratic equation are clear and make perfect sense, then this suddenly appears:

"Higher-degree equations may be quadratic in form, such as:

2x6th power + 3xthird power + 5 = 0

Note that the highest exponent is twice the value of the exponent of the middle term. This equation may be resolved directly or with a simple substitution, using the methods that are available for the quadratic, such as factoring (also called factorising), the quadratic formula, or completing the square."

Bringing in a non-second degree equation is a good idea, but only as an example against the previous and following equations. The note confuses me. Is it my imagination, or has that +5 also made this equation devilishly difficult? Also, could I see just exactly how one would solve "This equation..." especially with the quadratic formula?

I think what the author had in mind was to introduce another variable

u=x^{3}

. Then the equation becomes

2u^{2}+3u+5=0

which is quadratic in

u

. We could use the quadratic formula to solve for

u

and then get

x

since

x=u^{1/3}

. 68.123.46.200 03:04, 18 January 2006 (UTC)

Above explanation is perfect and should be in the orginal text. So the text might be amended to read:

You can solve this non-second degree equation directly, or if you choose not to solve it directly, because the highest exponent is twice the value of the exponent of the middle term, you may instead substitue a variable for $x^{2}$ , say "u", and use quadratic methods, such as factoring (factorizing),the quadratic formula, or completing the square to solve for $2u^{2}+3u+5=0$ .

Simplifed way?

i'm in gr 9 and i'm doing gr. 10 math. could u guys who work on all these math related articles have a simplified version in a small corner for those who want to get ahead but need to get a more simple explanation, not a year's worth of knowledge in a short paragraph?

C++ program

The following C++ program computes the quadratic formula:

//quadratic formula
//(C) 2006 42istheanswer
//Released under the GNU GPL or GFDL.
#include <iostream>
#include <cmath>
using namespace std;
int qformula(long double a,long double b,long double c,long double *output){
long double answer1;
long double answer2;
answer1=(-b+sqrt(b*b-4*a*c))/(2*a);
answer2=(-b-sqrt(b*b-4*a*c))/(2*a);
output[0]=answer1;
output[1]=answer2;
}
int main(){
long double a,b,c;
long double answer[2];
cout<<"A:";
cin>>a;
cout<<"B:";
cin>>b;
cout<<"C:";
cin>>c;
cout<<"Solving...\n";
qformula(a,b,c,answer);
cout<<"Answer 1 is:"<<answer[0]<<"\n";
cout<<"Answer 2 is:"<<answer[1]<<"\n";
return 0;
}

This is just a simple example, as it does not make sure that a is not zero.

The fact that dformula is defined for int outputs but the answer is outputted by reference would cause errors in some compilers. i would suggest either replacing "int" in the qformula definition line with "void," or replacing it with "long double" and passing the output through "return output;". -- He Who Is^{[ Talk ]} 22:41, 25 July 2006 (UTC)

a=0???

Why does a have to = 0? I'm pretty sure it doesn't have to. One can use the quadratic formula with ANY expression with the form $ax^{2}+bx+c/$ , regardless if a is 0 or if a, b, and c are 0. It just becomes more work than neccessary in those cases. I think we should remove the condition that a=0. Fresheneesz 04:50, 12 May 2006 (UTC)

The condition is "a is NOT equal to zero". In the formula given, the roots are not defined if a = 0 as you are dividing by zero. Of course if a = 0, then bx + c = 0 and x = -c/b. There is only one root. --Bduke 05:15, 12 May 2006 (UTC)

I'm sorry, i meant "why does a have to NOT = 0" - my fault. I see what you're saying, my mistake again. But thats a condition put on the quadratic formula, not on quadratic equations - however the condition is also put on a quadratic equation (at the top of the article), thats what i'm trying to take issue with. Fresheneesz 05:25, 12 May 2006 (UTC)

Well, if a = 0, you do not have a quadratic equation. i.e. a polynomial of order 2. You have a linear equation - a polynomial of order 1. --Bduke 05:30, 12 May 2006 (UTC)

I see your logic, but I'm a big fan of generaliation. Instead of using forms for solving first order DEs, I'd rather use the method for solving 2nd Order DEs, as they require me to remember less - and they still work even for first order DEs.

I was also wondering, since the quadratic formula doesn't work when a = 0, does this mean that the quadratic formula is an *approximation* whose error becomes greater and greater as a approaches 0? I just did some calculations, with the equation

ax^{2}+3x+2=0\

. I found that when a is 1 the quadratic formula works perfectly - but when a is .00000000000001, then at the supposed zero

ax^{2}+3x+2=.00159856

rather than equaling 0. I also found that if a is not 1, the quadratic formula also doesn't work very perfectly - but may give complex number with non-zero magnitude. This leads me to believe that the quadratic formula is an approximation. Fresheneesz 08:32, 14 May 2006 (UTC)

Perhaps it has something to do with the "loss of significance" ?? Fresheneesz 08:36, 14 May 2006 (UTC)

If you try to use the quadratic formula with a = 0, you get division by zero. A linear equation should not be thought of as a quadratic equation with a =0 ; they are different beasts. -lethe ^{talk +} 10:42, 14 May 2006 (UTC)

I think that just such thinking is healthy. Not only does it allow a person to notice the transition between -a, a line, and +a, but it forces the thinker to think about the realities of handling 0 in real mathematics. People avoid the odities of zero whenever possible, but it is a fact of life that is sometimes harder to reconcile than imaginary numbers. People aren't knowlegable enough about 0, I know I'm not. Fresheneesz 19:44, 14 May 2006 (UTC)

I don't understand. The quadratic equation simply doesn't work with a = 0. I don't know any way of thinking that will save you. -lethe ^{talk +} 19:55, 14 May 2006 (UTC)

I can accept that. But I would guess that the quadratic equation also doesn't work very well when a is very close to 0. Fresheneesz 18:36, 15 May 2006 (UTC)

Not at all. It's 100% accurate unless a = 0. Any errors should be accredited a person or computer's inability to handle such small nubers. Quadratic Equation#derivation gives a proof that a quadratic equation is logically identical to the quadratic formula.

Yea if a = 0 you do not have a quadratic equation, and if you try it in the formula you'll get division by zero which equals infinite. As long as a > 0 or a < 0 it is 100% correct no matter what. - T. Stokke

It does not equal "infinite". Division by zero is undefined full stop. Impossible. Moreover, infinity is not a number. —The preceding unsigned comment was added by 203.118.142.253 (talk) 09:34, 11 April 2007 (UTC).

Well, If a were to equal 0, then it would cancel out the ax^2 part, which would make the equation y= bx + c, which makes it a linear equation. —Preceding unsigned comment added by 96.233.5.37 (talk) 20:56, 13 January 2008 (UTC)

Derivation of quadratic formula

I really do think that Fresheneesz should have discussed splitting off the derivation to another page here before doing it. What do others think? --Bduke 06:12, 12 May 2006 (UTC)

Sorry, I was being bold. Derivations are useful - only if you're looking for them. Do you think it was a bad idea? Fresheneesz 08:21, 14 May 2006 (UTC)

I think it is unnecessary spliting stuff that is best kept together unless the article is getting too long, but I'm not going to get worked up about it. Let's leave it and see whether others agree or disagree. --Bduke 10:40, 14 May 2006 (UTC)

Aparently User:Alex_Bakharev disagrees with me. He put a merger notice on that page, but didn't disucuss why. I still think that long derivations don't belong on main pages. Fresheneesz 23:49, 29 May 2006 (UTC)

I have added the mergefrom tag to this article. The tags should go as pair. I may join the debate later. --Bduke 00:32, 30 May 2006 (UTC)

"generalized" form vs standard form

The definition at the top says the "generlized form is "ax^2+bx+c = 0 where a does not = 0" - but I think this is *standard* form. General form should be "y=ax^2+b+c" (or "f(x)="). Comments? Fresheneesz 08:39, 14 May 2006 (UTC)

I agree. There is nothing generalized about the form. Perhaps the author was an eightgh grader who thought that using variables instead of numbers for the coefficients makes it "generalized". I don't agree with that reasoning though. I've changed it to say "standard form". -lethe ^{talk +} 11:03, 14 May 2006 (UTC)

Presumably the author meant to say: "general form", as in contrast to specific forms of the QE like x^2+x-1 = 0. I think this is "standard" use: [1], [2], also for other equations on Wikipedia: Linear equation, Recurrence relation, Navier-Stokes equations and many more. --Lambiam ^Talk 18:26, 2 June 2006 (UTC)

Yep, that sounds about right. I've changed it. -lethe ^{talk +} 19:39, 2 June 2006 (UTC)

If its not standard, it is *definately* in no way general. I'm going to remove that modifier completely for now. Fresheneesz 19:07, 4 June 2006 (UTC)

Minus plus sign

"Note also that the signs distinguishing the two roots are reversed."

What does this mean? Whats the significance of a minus-plus sign vs a plus-minus sign? On Plus-minus_sign, it says the minus-plus sign is used to mean that the minus-plus sign is *minus* when the plus-minus sign is *plus*, and vice versa. From what I could understand, minus-plus only takes on significant meaning when used *with* a plus-minus sign (in the same expression or equation). Fresheneesz 08:46, 14 May 2006 (UTC)

The minus/plus has meaning when compared to the plus/minus of the other expression. -lethe ^{talk +} 11:03, 14 May 2006 (UTC)

Yes, however the quadratic formula is not compared to any other expressions on this page. The equation could use a plus-minus sign until it is compared with another equation involving an opposite plus-minus sign. But I do think its a good thing to introduce here, and so I'll just add a note about it saying something like: "The minus-plus sign only takes on a different meaning than the plus-minus sign when it is used with a plus-minus sign - see plus-minus sign for more detail."

Well, just remember that the minus/plus in that expression does have a meaning. Don't change the text to make it seem like it doesn't. -lethe ^{talk +} 19:41, 14 May 2006 (UTC)

Yea, I understand its significance now. Do you think that text is misleading? How would you reword it? I do think that some note to that effect is neccessary. Fresheneesz 19:46, 14 May 2006 (UTC)

I changed the sentence to be more clear (I hope). Fresheneesz 19:48, 14 May 2006 (UTC)

I've changed it even more. I think maybe it's best to just say exactly what the symbol means, which should make it clear the reason for using the opposite symbol here. What do you think of my version? -lethe ^{talk +} 19:54, 14 May 2006 (UTC)

Well, technically, it doesn't matter in what order you calculate the solutions, as long as the minus plus sign is always the opposite of the plus minus sign. I'm not saying your version is worse, but I wouldn't have understood it. I think of plus minus signs as short hand notation for two different solutions (an unordered pair). Fresheneesz 07:00, 15 May 2006 (UTC)

If you consider the solution set to be unordered, then there can be no distinction between the two roots, and the plus/minus minus/plus distinction has no meaning. But of course, this solution set is ordered, as indicated by the subscripts. I've an idea, which may make the idea even clearer. Tell me what you think. -lethe ^{talk +} 10:17, 15 May 2006 (UTC)

"If you consider the solution set to be unordered, then ... the plus/minus minus/plus distinction has no meaning" - not true, the solution set can be unordered and the disctinction between the two is just as important. For example,

1\pm 2\mp 3

equals 0 or 2, but its not an ordered pair. It doesn't matter if you get 2 then 0, or 0 then 2. In the quadratic equation, the roots aren't the "first" and "second", even though some might call them that - first vs second has no significance in any mathematical way for solutions sets.

"of course, this solution set is ordered" - I don't know why the ordering would matter, as long as the plus-minus sign is always calculated as the opposite of minus plus. Fresheneesz 18:34, 15 May 2006 (UTC)

As sets,

2\pm 1

and

2\mp 1

are the same. Both sets are

\{1,3\}

. As ordered sets, the first is

(3,1)

and the second is

(1,3)

. As you see, they are different. -lethe ^{talk +} 19:20, 15 May 2006 (UTC)

If they *were* ordered sets - yes I can see how they're different. But I don't see how they are in any way ordered, except by your arbitrary convention. What exactly does this ordering *mean* about those sets (which, if i'm not mistaken, are by definition unordered lists, as it says in the article on sets, "Set identity does not depend on the order in which the elements are listed"). Fresheneesz 21:44, 15 May 2006 (UTC)

Put a different way - how can you say one integer root for 4 is "first" and another "second" ? Is -2 first? or is 2 first? My opinion is that they aren't ordered solutions. Fresheneesz 21:46, 15 May 2006 (UTC)

This isn't a matter of opinion. It's a matter of notation. Either you want to denote the solution set as ordered, or you don't. The notation used in the article indicates ordered solution sets. This has the benefit that you have a correspondence between the two forms of the solution. -lethe ^{talk +} 22:12, 15 May 2006 (UTC)

Sets are not ordered. That is part of a definition of a set. You can *think* of a plus-minus sign as having an "order", but this only has to do with the way you learned it - it does not have to do with the reality of the math. The only thing that matters when calculating something like

a\pm b\mp c

is that the plus-minus sign is always rendered as the opposite of the minus-plus. It has nothing to do with order. a-b+c is not first nor second, and a+b-c is also neither. They are both solutions of equal importance. Fresheneesz 19:13, 17 May 2006 (UTC)

It's true that sets are not ordered. Ordered sets (also called tuples) are ordered, of course. The notation in the article denotes ordered sets (that's what subscripts do for you. The solution set x₁ = 3, x₂ = 5 is distinct fromx₁ = 5, x₂ = 3 because of the subscripts, even though the two unordered sets coincide. So let me just say this again: the notation used in the article denotes ordered sets. With the order, the two plus/minus signs are distinct). Now, I'm not really interested in being taught set theory by you. I think you seem to have a fundamental misunderstanding, but I don't know how to correct it without repeating myself. Do you still have a question about why the minus/plus sign is distinct? Because this conversation seems to be going in circles. -lethe ^{talk +} 19:29, 17 May 2006 (UTC)

I'm just not convinced that the convention of "ordered sets" is standard or most used. In fact, could I see some sort of references for such thinking? Also, could you show me exactly what meaning of "ordered set" you're talking about, because there are different meanings. Fresheneesz 00:38, 19 May 2006 (UTC)

I don't know off the top of my head any reference which explains the notational significance of subscripts on variables. I think it's a very basic notational concept. I assure you that the use of subscripts is quite standard, and so is the fact that subscripts entail ordered sets. I don't know what "different meanings" you could be referring to. There is only one possible meaning that one can assign to a collection of indexed variables.-lethe ^{talk +} 00:57, 19 May 2006 (UTC)

Check out ordered set for a couple different meanings. I tried looking up ordered set, and none of them mention the type you're talking about. Fresheneesz 03:24, 19 May 2006 (UTC)

I don't see anything in that article relevant to this discussion. -lethe ^{talk +} 04:43, 19 May 2006 (UTC)

Thats my point. Whatever "ordered set" you're talking about doesn't seem to exist whenever I try to look for it. Fresheneesz 22:04, 19 May 2006 (UTC)

So wait, are you trying to show me meanings of the term "ordered set", or are you asking for me to show you something? Your first comment "check out ordered set" makes me think the former, while your latter comment "no meaning seems to exist" makes me think the latter. -lethe ^{talk +} 08:39, 20 May 2006 (UTC)

Its both. I showed you some definitions, and I'm asking what definition *you're* talking about. Fresheneesz 20:21, 20 May 2006 (UTC)

Well, I'd point you to the article subscript, but it doesn't say anything at all. Instead I'll suggest that you check out some introductory math books from the library which explain what different notations mean. I don't have any particular reference. -lethe ^{talk +} 21:12, 20 May 2006 (UTC)

Ok... i'm not going to go to the library. Something so "basic" would surely be mentioned *somewhere* online. I don't have much of a problem with the notation though. I figured that the 1 and 2 meant that there were two solutions to the equation - but given that such notation seems to be extremely non-standard, I think it wouldn't be a bad idea to remove that notation. Along with the fact that the term "ordered sets" is never used to refer to the order in which its members exist, and not only that - sets are never ordered - I'm going to change the plus-minus thing to not imply ordered solution sets.

I hope you think thats logical too, because it really does seem like ordered solution sets aren't mentioned anywhere, and aren't implied by the minus-plus sign. If such notation and thinking was very common, I would have no problem agreeing with you - but it seems as though the notation isn't common, and probably won't be very well understood by people refering to this page. Also, I think it would mislead students in thinking that sets are ordered. Fresheneesz 11:06, 21 May 2006 (UTC)

You're wrong that this notation isn't very common, and you're wrong that it is not standard. It's OK to correct articles that are not clear, but it is not OK to rewrite articles to address your own fundamental lack of understanding. Your failure to understand what subscripts mean is quite atypical. -lethe ^{talk +} 13:26, 21 May 2006 (UTC)

I know what subscripts mean, don't insult me. I would be happy to admit that I'm wrong if you show that the notation is

common,
denotes unordered sets,
that unordered sets mean what you say they mean.

And once again, the subscripts are not a problem for me - only the "ordered set" idea. Fresheneesz 19:06, 21 May 2006 (UTC)

I gave an example earlier in this thread of how subscripts entail ordered sets. This happens every time you use subscripts in your notation; it's what the subscripts mean. So I guess you are not disputing that subscripts are common, but rather you're disputing that subscripts imply an ordering of the values. To that, I can only respond: you should think about the example a little. -lethe ^{talk +} 19:23, 21 May 2006 (UTC)

Look, thats not "what subscripts mean". Subscripts are used to further specifiy a variable, and double subscripts are used to refer to something between two points. Subscripts are simply part of a variable to help further explain the meaning of that variable. I've seriously never seen subscripts to denote two ordered values of a set. I understood the example, (I assume you're talking about "...solution set x₁ = 3, x₂ = 5 is distinct fromx₁ = 5, x₂ = 3 because of..."), but it doesn't prove that thats the way solution sets are thought of. Also, I went back and looked at the page on tuples, which are apparently not ever called "ordered sets", but are called "ordered lists".

Please, I don't want this to turn into a fight. I'm just asking for some sort of evidence that what you say is true. Thats how wikipedia works. One of the main policies at wikipedia is verifiability, and this subscript notation lacks that without some sort of evidence that its used in that way elsewhere. Fresheneesz 20:10, 21 May 2006 (UTC)

I think the subscripts are being used to point out a certain relation between the two equations. A cookie for anyone who tells me what the relation is. :D -- 127.*.*.1 20:29, 21 May 2006 (UTC)

It certainly doesn't prove that "that is the way solution sets are thought of". Indeed, solution sets are usually not thought of that way. But as I have said, it is a matter of notation. The original author of this page chose to label the solutions with subscripts. It is, as I have said, merely a matter of notation. -lethe ^{talk +} 00:23, 22 May 2006 (UTC)

Ok. However, my argument is that that notation is non standard. Don't get me wrong, I fully see how the idea of ordered solutions can be used, but I don't find it to be either a useful idea or a common idea. My whole problem with it is that it seems to simply not exist. Why is this idea that you admit is non-standard so important to keep on this page? I like good non-standard ideas as much as the next crackpot, but I don't see the use of using ordered sets when the minus-plus sign is much more easily explained as "the opposite of the plus-minus sign", plain and simple. Fresheneesz 06:10, 22 May 2006 (UTC)

Well, I am going to continue replying to you, even though everything I say here will be something I've already said in this thread. We've fallen into the pattern of the repetitive circular argument which I was hoping to avoid. Perhaps if I discuss my points at greater length, then you'll be more willing to consider them? So the notation used on this page has an obvious advantage. It allows you to have a correspondence between the roots written in the form

x_{\pm }={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

and the form

x_{\pm }={\frac {2c}{-b\mp {\sqrt {b^{2}-4ac\ }}}}

if these two solution sets were unordered, then there would be no way to denote which of the two roots from the first solution set was which root from the second solution set. But with subscripts, we have the two equations

{\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}=x_{+}={\frac {2c}{-b-{\sqrt {b^{2}-4ac\ }}}}

and

{\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}=x_{-}={\frac {2c}{-b+{\sqrt {b^{2}-4ac\ }}}}

It is my opinion that seeing the correspondence between these two equations is useful. Knowing which root is which when switching between the two methods for calculating roots is useful. This cannot be done without some ordering of the two roots. So I disagree with your assessment that the idea isn't useful. This is the point which I made at the top of this thread with the comment "This has the benefit that you have a correspondence between the two forms of the solution."

I guess lethe gets the cookie... -- 127.*.*.1 15:43, 22 May 2006 (UTC)

As for your claim that this is nonstandard; well it's my opinion that the notational device of ordering elements of a set by denoting them with subscripted variables is very common. It might be a little odd to force an ordering on a set which has no natural order, but I think it's useful in this context. Here at wikipedia, we are allowed to make our own choices of notation, this does not violate our original research policy. As I have been saying all along, this is a matter of notation. If a notation is useful to illustrate a point, then let's use it. Changes of notation are trivial enough that we do not have to worry about them being standard or not. The obvious exception is of course new notation. We cannot invent new notation here at wikipedia, but of course subscripts are not new notation; they are one of the most common notations in mathematics. -lethe ^{talk +} 15:38, 22 May 2006 (UTC)

Like you've said, you've said all this before. I read, and understood, everything you wrote. Rehashing the info doesn't convince me any further. I *do* understand how the ordered set idea can be used. I *do* realize what you're saying about this.

"This cannot be done without some ordering of the two roots." - As for that quote, I've already replied saying that:

"The only thing that matters when calculating something like $a\pm b\mp c$ is that the plus-minus sign is always rendered as the opposite of the minus-plus". One doesn't need order to do that. And you haven't really addressed the issue of verifiability. Obviously subscripts are common - once again I really don't care about the subscripts - but bringing in the idea of an "ordered set" into the picture makes me wonder where that comes from, and you have yet to show me (ie sources).

In any case, you're right that this is going nowhere. I propose we consolidate our views and points under a new header, and wait for a consensus to appear. Does that sound fair? Fresheneesz 17:08, 22 May 2006 (UTC)

Fine with me. I suppose no one is going to want to wade through the novel written above. -lethe ^{talk +} 17:18, 22 May 2006 (UTC)

first root, second root?

The discussion above has gotten very long, and me and lethe want to come to a consensus on the issue. I'll outline a list of points that were brought up above so noone has to read that junk:

Lethe asserts that the plus-minus and minus-plus signs ( $\pm$ and $\mp$ ) order the roots in the two different versions of the quadratic formula. Thus there would be a specific "first" root, and a specific "second" root - a concept lethe described as an "ordered set".
I argued that there cannot be a distinct "first" or "second" root - as solution sets (and all sets) are non-ordered in that sense.
Lethe then that the "ordered set" idea was implied in the subscript notation.
I then argued that the minus-plus sign does not need a concept of "ordered sets" to be used, and furthermore that I could not find any sources available that could verify the concept of an "ordered set".
Also, although lethe never used the words "ordered set" in the article, I hope that will stay that way, because I think the idea of an "ordered set" would mislead readers into thinking that members in a set can have an order by which they're listed.

I've probably left some things out, so I let you fill in the gaps lethe. Fresheneesz 02:58, 23 May 2006 (UTC)

The plus/minus signs do not order anything. The subscripts do. Because of this, the use of a

\mp

has a distinct meaning from the

\pm

sign. The meaning is given by two equations above. -lethe ^{talk +} 03:15, 23 May 2006 (UTC)

Not to grow this into another long string, but the minus-plus sign is distinct from the plus-minus sign whether the subscripts are there or not. And if the subscripts are where this "ordered set" idea comes from, then isn't it misleading to write that "The minus-plus sign indicates that the first root (as defined above) is calculated with the minus sign and the second root with the plus sign" - thus indicating ordered sets? If the subscripts are where the ordered set thing comes from, then why say that the plus-minus sign inherintely incorporates the idea of an "ordered set"? Fresheneesz 07:48, 23 May 2006 (UTC)

then why say that the plus-minus sign inherintely incorporates the idea of an "ordered set"? Luckily I did not say that. the minus-plus sign is distinct from the plus-minus sign whether the subscripts are there or not. Oh? So what is the distinction between the expression

a\pm b

and

a\mp b

? If I don't denote with subscripts, those expressions are probably the same. -lethe ^{talk +} 12:34, 23 May 2006 (UTC)

Sure I agree. Which gives me some insight on a possible compromise. Why don't we discuss the parallel between the roots using the two different quadratic formulas. Instead of using the "ordered set" idea, why don't we simply say that the root in the form

{\frac {-b+{\sqrt {\Delta }}}{2a}}

(from the regular quadratic formula) is the same root as the root in the form

{\frac {2c}{-b-{\sqrt {\Delta }}}}

and vice versa? Fresheneesz 02:36, 24 May 2006 (UTC)

Removing the notation is one solution, but I don't think removing notations you don't understand is the right solution. Actually, you've said that you do understand, so I don't know why we're still discussing this. The article is fine the way it is. If you still feel you have to change it, I think you'll have to get some more people over to your side. -lethe ^{talk +} 03:08, 24 May 2006 (UTC)

I'm not sure I understand exactly what this argument is about. However I don't think there is any problem with the notation currently being used in the article. Fresheneesz, if you would like to propose a specific alternative notation, please do so here on the talk page, and I (and perhaps other editors) can express our opinions as to whether it would improve the article or not. Paul August ☎ 03:45, 24 May 2006 (UTC)

Heh, you know what's funny? I'm not sure what the disagreement is about either, and I've been up to my neck in it for like a week now. Fresh doesn't like the minus/plus sign, but I'm still not sure why not. And he really latched on to my dirty explanation in terms of ordered solution sets. -lethe ^{talk +} 04:17, 24 May 2006 (UTC)

Haha, this is such a ridiculous argument! In any case, my only problem with the article at the moment is the sentence "The minus-plus sign indicates that the first root (as defined above) is calculated with the minus sign and the second root with the plus sign"". My point is simply that talking about "first" and "second" roots could leave a reader wondering why the first one comes first, and why the second one comes second. This could lead to lots of questions about solution sets for those of us that would come here knowing about sets. For example, me and lethe both agree that sets are not ordered, and therein lies my beef: first and second roots imply some sort of order in the set.

The notation is perfectly fine with me. Fresheneesz 17:49, 24 May 2006 (UTC)

I don't think that this argument is ridiculous at all. To me, $x_{\pm }$ implies that the order of the two roots is dependent on the plus-or-minus sign in front of the radical under which the discriminant is situated - i.e., $x_{+}$ will be larger than $x_{-}$ (or the former will be positive, while the latter negative) when the discriminant is above zero. While it is true in some cases, it is false in others (it depends on various things). Personally, I'd say remove the subscripted plus-or-minus sign to avoid confusion. While others might not grasp the same implications as me, it's bound to cause confusion - and in the end, what purpose does it serve? The plus-or-minus sign in front of ${\sqrt {\Delta }}$ clearly indicates that two roots exist. What do others think?

As for the second form of the quadratic formula, I agree with what Fresheneesz said about saying that ${\frac {-b+{\sqrt {\Delta }}}{2a}}$ (from the regular quadratic formula) is the same root as the root in the form ${\frac {2c}{-b-{\sqrt {\Delta }}}}$ and vice versa, etc. Though, it should be said without the current weird notation. Sure, somebody who is very familiar with quadratics will understand the current notation, yet why do we want to confuse people who are less familiar with them?

— Absolutely Insane 07:59, 12 July 2006 (UTC)

I agree with Fresh that the words "first" and "second," with respect to the roots, it somewhat ambiguous, so I propose it merely be changed to: "The minus-plus sign indicates that the lesser root is calculated with the minus sign and the greater root with the plus sign"" or something to that effect. While not nessecarily true with complex roots, I think it gets the point accross without getting too complicated. -- He Who Is^{[ Talk ]} 18:04, 12 July 2006 (UTC)

The "first" and "second" language is no longer in the article, and hasn't been for some time. Paul August ☎ 20:19, 12 July 2006 (UTC)

characteristic 2

The article says the formula works over any field of characteristic ≠ 2. So how do you solve quadratics over fields of characteristic 2? OK, I know that some quadratics over F₂ are irreducible. But what about the algebraic closure? -lethe ^{talk +} 05:24, 24 May 2006 (UTC)

I have added a section on solving the quadratic equation in characteristic 2. Comments welcome. -lethe ^{talk +} 12:31, 5 June 2006 (UTC)

There isn't much of a formula though. You have to find the lion's share of it by inspection, apparently. -lethe ^{talk +} 21:58, 5 June 2006 (UTC)

lists vs paragraph

I would like to write the quadratic formula in the form:

x_{\pm }={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

where x can have two possible values to solve the equation:

$x_{+}\$ is the x where the $\pm$ is interpreted as a $+\$ , and
$x_{-}\$ is the x where the $\pm$ is interpreted as a $-\$ .

Since lethe doesn't like listing like this, I can't change it on the page without consensus. I'll leave this comment here until people say whether or not they like it. Fresheneesz 11:40, 28 May 2006 (UTC)

In addition to your policy of using bulleted lists every time you want to explain a variable, I think your phrase "x_+ is the x with the ± interpreted as a +" is badly phrased. -lethe ^{talk +} 13:21, 28 May 2006 (UTC)

Once again, change the phrase, don't remove it. What about "x+ is the solution where the \pm is interpreted as a +" etc. Is that any better? What do you find wrong with the phrase? Fresheneesz 19:54, 28 May 2006 (UTC)

Besides being poorly phrased it doesn't add any thing. Paul August ☎ 21:07, 28 May 2006 (UTC)

How about:

x_{\pm }={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

where x can have two possible values:

x₊ when ± is interpreted as +
x₋ when ± is interpreted as −

It's a bit more concise, don't you think? If you think it's obvious it's easy to overlook, if you don't, you get the same information. I think this way might be clearer too. Shinobu 21:51, 7 June 2006 (UTC)

Well putting it like:

x_{\pm }={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

Would be incorrect, basically because x± does not equal that equation, the more logical approach would be x_1,2 (or) x₁ and x₂ =

{\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

—The preceding unsigned comment was added by T.Stokke (talk • contribs) 15:19, 1 January 2007 (UTC).

merge of derivations with this page

The articles have been merged without debate. I disagree with them, and I'm gonna have to revert this until further discussion is had. I disagree with merge. Fresheneesz 19:03, 4 June 2006 (UTC)

To be more specific, I disagree with merge because the derivations are long, and most readers would not want to see them. I think that enough is reason to separate them. Articles are made to be an efficient communication of information, not a list of everything that has to do with a topic. I simply think that the derivation bogs down the main article. Fresheneesz 19:25, 4 June 2006 (UTC)

So you did the split without discussing it, and someone merged it back, and you use their lack of discussion as an excuse to undo the merge? Perhaps we should discuss your split first. I don't like the split. -lethe ^{talk +} 19:52, 4 June 2006 (UTC)

I like having the derivation here — I didn't agree with the original split — however I wouldn't object to it coming further down in the article. I also don't think the "alternative form" should be as prominently located as it is, and it could be moved further down as well. Paul August ☎ 21:05, 4 June 2006 (UTC)

Lethe, can you please be more civil in discussions, and not intentionally try to insult/ridicule me? I split without discussion, this is true. But now that I did the split, there is now discussion about it on this talk page - fairly recent discussion at that. And after that discussion there is a very clear no consensus. Reverting the split without providing explanation or having discussion when there is previous discussion about it is improper. THAT is my argument lethe.

Btw, could you please provide a *reason* why you "don't like the split". Fresheneesz 22:30, 4 June 2006 (UTC)

Yes, thank you for the reminder. I will try to be more civil. I did mean my above comment to insinuate that you are being hypocritical, and for that I do apologize, I was out of line. Now on to the matter at hand: the split was inappropriate, and the merge is supported. I will continue to revert edits that I think are bad. Revert wars are viewed as counterproductive, but I'm conservative about math articles, if you want to make a big change, you should discuss it first, or else there is a good chance that I will revert. In the mean time, you should also view this AfD for a list of people who support the merge. -lethe ^{talk +} 22:37, 4 June 2006 (UTC)

The split has been active for some time now, and I think it would have been proper to discuss the merge before merging. But whatever. In any case, that AfD is not a discussion or consensus about whether to merge. It is a discussion about deleting the page. In fact, most people don't even mention a merge. You and Bduke are the only ones in support of the merger on that AfD. However, User:Alex Bakharev has put a merge notice on the derivation page, so I suppose he supports merging. The guy that merged it supports it, you support it, Bduke supports it. So its 4 on 1. And Avenue has no opinion either way.

That being said, I still want to know why you think the derivation should be on this page. I really think its clutter most readers wouldn't want to read. I thought it was pretty elegant to have an easy link to it right next to the equation. That way readers would know we have a derivation after looking at the formula, but they wouldn't have to sort through it on this page if they didn't want to (which I think describes most people comming to this page). Remember, this is a page about a begginers math subject. Fresheneesz 00:45, 5 June 2006 (UTC)

I think you've missed a lot of others who've voiced opinions on this matter. -lethe ^{talk +} 01:13, 5 June 2006 (UTC)

Fresheneeze, you seem not to have taken into account my opinion, expressed above. If there is going to be a derivation at all, then it certainly belongs here in this article. There is certainly room for it. And as far as being "clutter", as I said above, moving it down in the article might help. Paul August ☎ 01:28, 5 June 2006 (UTC)

I have moved it to the bottom. -lethe ^{talk +} 01:37, 5 June 2006 (UTC)

Sorry paul, didn't mean to leave you out. Looks fine at the bottom. Fresheneesz 08:45, 5 June 2006 (UTC)

New version

I've revised the "Quadratic formula" and "Discriminant" sections, reorganized some of the content, added three new sections, as well as making other less major changes elsewhere. I thought it might be a help to other editors if I detailed the more significant changes here:

Quadratic formula:

Eliminated the clever but novel (and perhaps confusing) use of $\pm$ as a subscript, and the need to use and explain the $\mp$ , sign.
Moved the discussion of the "alternative formula" to its own section: "Alternative formula", further down in the article.

Discriminant:

Removed the introduction (and later use) of the symbol: $\Delta .\,\!$
Added the necessary restriction to real coefficients.
Changed the order of consideration of the three cases.
Added the necessary requirement of integer coefficients, to the statement that the discriminant being a perfect square implies rational roots.
Eliminated the redundant specification of the roots for the positive discriminant.
Removed the discussion of x-intercepts to its own section: "Geometry".
Removed the discussion of factorization to its own section: "Quadratic factorization".

Geometry:

Moved the content from the previous "Discriminant" section, on the relationship between the discriminant of the quadratic equation and the graph of the corresponding quadratic function, to this new section.
Adding the necessary qualification about the domain and codomain of f.
Expanded and clarified (I hope).

Quadratic factorization:

Moved the content on factorization from the "Discriminant" section, to this new section.
Expanded and clarified (I hope).

Application to higher-degree equations:

Changed the title of this section from "Solving equations of a higher degree".

Derivation:

Moved the derivation of the "alternative formula" to the new section which follows.

Alternative formula:

Moved this content from the "Quadratic formula", and "Derivation" sections.

I also reordered the sections, to what to me made more sense. I welcome comments or critisisms of any of this, and also hope other editors will verify and fix any of my mistakes.

Paul August ☎ 21:59, 24 July 2006 (UTC)

The square-root step

When completing the square and having the (x-a)^2 = b of the equation I think it is much more pedagogical to rewrite this as (x-a)^2-(sqrt(b))^2 = 0 and recognize the conjugate rule which yields (x-a+sqrt(b))(x-a-sqrt(b) = 0. In this form it is obvious that there are two solutions. The standard way of writing x-a = +- sqrt(b) confuses many students into believing wrong things like sqrt(4) = +-2 or sqrt(-1) = i.

But sqrt(-1) = i. And I don't see how there could be any confusion. Everyone knows that (-x)^2 = x^2. So if x^2 = b, x = +-sqr(b). Note also that (x-a)^2 = b is not exactly the same as (x-a)^2-(sqrt(b))^2 = 0. Your version also looks more complex (in the non-mathematical sense). Shinobu 23:26, 25 August 2006 (UTC)

No, sqrt(-1) is not equal to i! To define the square-root of negative numbers (and complex numbers), you need to specify the branch for the logarithm.

The way I was tought this, it is implicit in the notation with the root sign that you want the principal value. There are also authors who give the functions a different symbol from their multivalued counterparts. Dettman does this with the logarithm, so if you want to use a multivalued square root, you could rewrite it in terms of logarithms and use the multivalued symbol. Of course, this is just one perspective. You can also define the logarithm to be a singlevalued function on a more complex kind of surface, but I think that is well beyond the scope of this article. Shinobu 12:58, 13 September 2006 (UTC)

Alternative version

I've noticed several people here have asked for a simplified version, perhaps this would be of some use. In college, an engineering professor of mine, originally from Iran, would frequently use the following form of the equation. He said he learned it as the standard form in Iran and seemed suprised that none of us knew it, doubly so that we attempted to memorize the much longer and more complicated version.

First, you must rewrite the original polynomial into the form $x^{2}+2bx+c=0\,\!$ by dividing through by a and taking b to be half of the second coefficient. After that, the quadratic equation simplifies to:
$x=-b\pm {\sqrt {b^{2}-c}}\,\!$
Substituting $b=b'/(2a)\,\!$ and $c=c'/a\,\!$ yields the original quadratic formula. -Anþony 04:53, 1 September 2006 (UTC)

Thanks a lot

i just wanted to say thanks to everyone who put together this superb page! it has saved my accademic behind on a number of occasions! thanks for the useful info that is explained it simply enough for a very un-mathematical person like me! :) 207.200.116.198 22:30, 22 October 2006 (UTC)

Quadratic Equation vs. Quadratic fuction

Whats the difference? Is a quadratic function just a quadratic equation that is a function? —Preceding unsigned comment added by 65.96.69.184 (talk • contribs) 17:47, November 4, 2006

Equations and functions are two different kinds of things: see equation and function (mathematics). Here is a short explanation:

A quadratic expression is a polynomial in one variable in which the highest power of the variable is two. For example: 3r² + 5r.
A quadratic equation (in standard form) is the statement that a given quadratic expression is equal to zero. For example: 3r² + 5r = 0.
A quadratic function is a function that associates each element x in its domain with the value obtained by substituting x for the variable in a given quadratic expression. For example, the function that associates each x with the value 3x² + 5x.

Paul August ☎ 19:05, 4 November 2006 (UTC)

Formula for factoring

Don't know if it's relevant or not, but this formula made classwork a whole lot easier for me when I was asked to factor quadratic equations. :)

$\ ax^{2}+bx+c=(yx+{\frac {b+{\sqrt {b^{2}-4ac}}}{2z}})(zx+{\frac {b-{\sqrt {b^{2}-4ac}}}{2y}})\$ if $\ yz=a\$

Yanah 22:28, 9 November 2006 (UTC)

Quadratic Vs. 2nd degree equations

Hello, this is a language issue for me and i need to know in order to translate properly, is Quadratic an 2nd degree equations the same thing? is x²+y²=55 for example a quadratic (i dought!)? or does it only match the form Ax²+By+C=D (D=0 ?) ?, thanks.--192.116.17.51 16:24, 19 November 2006 (UTC)

{{aan}