Talk:Quadratic equation/Archive 4

Archive 1

Archive 2

Archive 3

Archive 4

Archive 5

Automated reformatting tools can break the equations

Plastikspork modified the page using some kind of automated tool on 28 February 2009. The comment made was: "Script-assisted whitespace removal, table & color simplification, and link repair." Perusing the diffs, I see several cases where the "simplifications" removed non-breaking spaces, which I consider of dubious benefit. But more importantly, several equations have been mangled, and no longer render at all; instead, we get a syntax error. Equations in the Derivation and Alternative formula sections are affected by this. I'm going to go in and fix the broken equations by copying them in from the previous version of the page.

In the future, it might behoove whoever edits a page of fundamental importance like this to check and make sure their edits didn't break something fundamental. This requires looking through the entire article. In this case, the breakage would have been easy to spot, since the system helpfully renders the broken math markup in bold red. 130.13.47.162 (talk) 07:30, 28 February 2009 (UTC) R. Poole

Thanks for alerting me to a bug in the script. I actually did check after running the script, but my browser caches the images, so I didn't spot it until I cleared my cache. Please do let me know if you notice any damage as I am continuously trying to refine the script. Thanks! Plastikspork (talk) 16:10, 28 February 2009 (UTC)

Mistake in the Derivation?

In the derivation of the quadratic formula, I think that the equation with the absolute values should be removed. With their removal, the derivation can be used for finding the roots of a quadratic equation with complex coefficients as well. Please check this out, I may have overlooked something. Thanks! (24.46.176.191 (talk) 02:05, 1 March 2009 (UTC))

4/8/2009 I am a frequent user of Wikipedia. Recently, at the top of the article "Quadratic Equation" was an offensive sexual statement. Looking for a way to report this, I ended up editing the article, simply deleting (by backspacing) the offending material. To my amazement, after clicking "Save page", the offensive statement was removed. The statement was absent when I reloaded the page several hours later.

I can't believe that just anyone, such as myself, could actually edit a Wikipedia article. If this is so, what is to keep some malicious individual from intentionally changing material within an article to make the article worthless, from the standpoint of accuracy? —Preceding unsigned comment added by 98.71.134.199 (talk) 14:07, 8 April 2009 (UTC)

See WP:VAND. ²³¹₉₁Pa (chat me!) 05:24, 16 December 2009 (UTC)

I placed the solution in the introduction

I came to this page, and was very surprised not to find the famous solution

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

in the introduction. In fact I had to scroll all the way to the bottom of the page to find it. Since the quadratic equation, and its solution, are one of the few things that many people recall from high-school algebra I think it is important to put it before any derivations. Is there any reason not to do this? mislih 15:39, 30 April 2009 (UTC)

For most of this articles existence the first section following the lead was called "Quadratic formula", It was recently deleted (don't know why) and I've just restored it. Paul August ☎ 16:28, 30 April 2009 (UTC)

sum & product

And why not to mention applications to quadratic systems like x+y=s, xy=p ? If they are already in some wiki article (but where?) a link shoul suffice. --84.221.208.46 (talk) 17:11, 20 May 2009 (UTC)

Simplified expression

Dividing the quadratic by a gives the simplified form of x² + px + q = 0, where p = b⁄a and q = c⁄a. This in turn simplifies the root and discriminant equations to

$x={\frac {1}{2}}\left(-p\pm {\sqrt {p^{2}-4q}}\,\right)$

$\Delta =p^{2}-4q$

This form appears in several popular books, e.g., John Derbyshire's Unknown Quantity: A Real And Imaginary History of Algebra (ISBN 0-309-09657-X). Is this worth mentioning in the article? — Loadmaster (talk) 20:16, 27 August 2009 (UTC)

Dividing through by a first is just converting to a monic polynomial? Plastikspork _―Œ^(talk) 21:01, 27 August 2009 (UTC)

Exactly. The question is, is the quadratic equation solution commonly presented in the reduced (monic) form often enough to justify being mentioned (as a variant solution, perhaps) in the article? — Loadmaster (talk) 22:25, 27 August 2009 (UTC)

New derivation by Eric Norby

The first definition give the "general" form of a quadratic.

It is something like ax^2+bx+c=0

I hate to pick beans but that is the equation for zero, with the general form for finding the roots of a quadratic.

Someone pointed it out earlier, and I agree.

That line should be changed to Y=ax^2+bx+c or f(x), at least.

It is very hard to plot a curve on an equation equal to zero.

If no one objects, I will make the correction myself.

Furthermore, I found the current "derivation" hard to follow, so I've written a much easier one to follow, providing for no typos, however, it is much longer. If anyone would like to use it to improve the current one, please be my guest. If no one objects, I will also change the current one to this easier flowing one below. If I get back here soon enough.

Note: This also derives the perfect square equation, and demonstrates how it works to solve the general form of the quadratic formula. I've also noticed the page for squaring is in need.

A perfect square is from the following squaring of two linear equations. (rX+s)*(rX+s) Multiplying them gets r^2X^2+2rsX+s^2 = (rX+s)^2 or (rX+s)^2=r^2X^2+2rsX+s^2 Note that it has no extra constants on the left side, just the squared term (rX+s)^2. A constant in this case is an arbitrary value that stays "constant" for any given equation. It is a variable here until the value for a specific equation is inserted. In this case, X is the only variable and all the rest are just constants, because they are not dependent on changes to x. They are constant relative to changes in X.

If we add a constant to one side we can add it to the other and the equations stay equal.

The goal is to derive something that will have extra constants on the left side, and still be equal to the general quadratic. Starting with an equation-form with extra constants on the left side, strive to find what the new constants could be.

In other words: Can extra constants k, m, and p, be found, such that the following is true. m(x+k)^2+p = ax^2+bx+c This is called the method of completing the square. (a,b, and c are given)

Let m=a and divide by m m/m(x+k)^2+p/m = (a/m)x^2+(b/m)x+c/m a/a(x+k)^2+p/a = (a/a)x^2+(b/a)x+c/a Canceling the a's (x+k)^2+p/a = x^2+(b/a)x+c/a This is where the provision for a not equaling zero comes in. Dividing by zero (a) causes infinities. An infinity in this case would mean the equation would cross the x-axis only once, and never again. The a=zero equation is called 'a line', when 'a' is zero, not a quadratic. Finding the root for 'a line' is much easier. Just divide a minus c by b.

Subtract c/a from both sides (x+k)^2+p/a- c/z = x^2+(b/a)x+c/a-c/a (x+k)^2+p/a -c/a = x^2+(b/a)x

Adding (b/(2a))^2 to both sides (x+k)^2+p/a -c/a+ (b/(2a))^2 = x^2+(b/a)x + (b/(2a))^2

The right side is now a perfect square (x+b/(2a))^2 (x+k)^2+p/a -c/a+ (b/(2a))^2 = (x+b/(2a))^2 Right side = perfect square

Choosing k = b/(2a) from noted equivalence (x+k)^2+p/a -c/a+ (b/(2a))^2 = (x+b/(2a))^2 (x+b/(2a))^2+p/a -c/a+ (b/(2a))^2 = (x+b/(2a))^2

Subtracting (x+b/(2a))^2 from both sides (x+b/(2a))^2-(x+b/(2a))^2+p/a -c/a+ (b/(2a))^2 = (x+b/(2a))^2-(x+b/(2a))^2 p/a-c/a+ (b/(2a))^2 = 0

Solving for p p/a-c/a+ (b/(2a))^2 = 0 p/a = c/a- (b/(2a))^2 p = c-b^2/(4a)

Checking our work m(x+k)^2+p Inserting our discovered values k=b/(2a), m=a, and p=c-b^2/(4a) Multiplying out (x+k)^2 in original equation m(x^2+2kx+k) + p = Substituting values a(x^2+2(b/(2a))x+(b/(2a)^2)) + c-b^2/(4a) = ax^2+2a(b/2a)x + a(b^2/(4a^2) + c-b^2/(4a) = ax^2 + bx + b^2/(4a) + c - b^2/(4a) = ax^2 + bx + c which is the equation of desired match, therefore the work has no errors.

Placing k=b/(2a), m=a, and p=c-b^2/(4a) into original equation, to get an equivalent equation m(x+k)^2+p = ax^2+bx+c a(x+b/(2a))^2 + c-b^2/4a = ax^2+bx+c

The quadratic formula is just finding roots (X axis zero crossings.) for I.e., when y = zero in (y=ax^2+bx+c) what is x. ax^2+bx+c = 0

Substituting in the equivalent equation a(x+b/(2a))^2 + c-b^2/4a = 0

Begin solving for x a(x+b/(2a))^2 + c-b^2/4a = 0 a(x+b/(2a))^2 = -c+b^2/4a (x+b/(2a))^2 = -c/a+b^2/(4a^2) (x+b/(2a))^2 = b^2/(4a^2)-c/a

Taking square root of each side Remember when you square a negative, it becomes a positive, so , when you that the square root of a square it is possible for the value to be positive or negative, thus the +/- symbol. sqrt(x+b/(2a))^2 = sqrt(b^2/(4a^2)-c/a) -/+(x+b/(2a)) = sqrt(b^2/(4a^2)-c/a) x+b/(2a) = +/-sqrt(b^2/(4a^2)-c/a) x = -b/(2a)+/-sqrt(b^2/(4a^2)-c/a)

Rearranging terms multiplying top and bottom of c/a with 4a^2 x = -b/(2a)+/-sqrt(b^2/(4a^2)-c/a) x = -b/(2a)+/-sqrt(b^2/(4a^2)-4ac/4a^2) x = -b/(2a)+/-sqrt((b^2-4ac)/(4a^2))

Pulling 4a^2 out of square root x = -b/(2a)+/-(sqrt(b^2-4ac))(2a)

Finally, solved for x x = (-b(+/-)sqrt(b^2-4ac))/(2a) This is the quadratic formula for finding roots.

I hope that this helps many of you. (There is no reason Mathematics should be obfuscated.) I submitting it here first to hopefully get some positive feedback, rather than just editing the main page.

Eric Norby (talk) 01:45, 7 September 2009 (UTC)

Just a few remarks:

• "but that is the equation for zero" => It is the "quadratic equation", which is the subject of the article.
• "That line should be changed to Y=ax^2+bx+c or f(x), at least." => The article title is "Quadratic equation". The expression you propose would pertain to a quadratic function.
• "... so I've written a much easier one to follow ..." => Please aquaint yourself with NOR.
• I have added a reference to the current derivation. Thanks for having drawn our attention to this.

DVdm (talk) 08:24, 7 September 2009 (UTC)

... and I cleared things up a bit. I don't think there's was a need to introduce these k and h constants. It's more conform to the specified source now. - DVdm (talk) 14:35, 8 September 2009 (UTC)

Use of quadratics?

Many, many, high schoolers and adults around the world, <gasp>, dont understand these equations. I think it would be helpful, since this is for a general audience, and these equations are part of every educated adults learning, even liberal arts types, to introduce the article with a little general explanation on why these equations are useful and what they can be used for. (Like calculating trajectory) To many people, it is not at all obvious. This article is very extensive and technical, which is good in one sense, but it reads as though it was written by mathematician *for* mathematician; which evidently it was (judging also by the discussion here). This article was delisted for a reason. Quadratic equations are probably one of the highest-searched-for basic-math terms on the web, especially by young students. The article, so to speak, should bear this in mind. Anyone agree? —Preceding unsigned comment added by 69.171.160.236 (talk) 00:33, 10 September 2009 (UTC)

alternate history

If the IRS had discovered the quadratic formula: http://www.cs.amherst.edu/~djv/irs.pdf

69.228.171.150 (talk) 01:43, 24 October 2009 (UTC)

Good one :-)

Alas, no room for this over here :-|

DVdm (talk) 10:15, 24 October 2009 (UTC)

Bad article

This article (the one here in Wikipedia) is as bad as it can be. Can not be understood by simple people who are not mathematicians and who have an interest of at least understanding how it works and why is it used for. I think it is not an article at all. It's just a mathematics class for scientists. Similar are other articles on scientific topics. Very disappointing. You think there is at least one place where you can learn things and it turns out it is a playground for mathematicians.

Ardi Kule —Preceding unsigned comment added by 80.78.69.234 (talk) 21:50, 5 December 2009 (UTC)

Babylonian quadratic

The article states that the Babylonians solved the equations $x+y=p,\ \ xy=q\$ as follows:

1. Form ${\frac {x+y}{2}}$
2. Form $\left({\frac {x+y}{2}}\right)^{2}$
3. Form $\left({\frac {x+y}{2}}\right)^{2}-xy$
4. Form ${\sqrt {\left({\frac {x+y}{2}}\right)^{2}-xy}}={\frac {x-y}{2}}$
5. Find $x,\ y$ by inspection of the values in (1) and (4).

As far as I am aware, ${\sqrt {\left({\frac {x+y}{2}}\right)^{2}-xy}}$ comes to ${\sqrt {\frac {x^{2}+y^{2}}{2}}}$, which does not equal ${\frac {x-y}{2}}$. In order to get that particular result, you need to subtract $2xy$ instead of $xy$.

J.Gowers (talk) 11:36, 22 December 2009 (UTC)

Please check your algebra:

${\sqrt {\left({\frac {x+y}{2}}\right)^{2}-xy}}={\sqrt {x^{2}/4+xy/2+y^{2}/4-xy}}={\sqrt {x^{2}/4-xy/2+y^{2}/4}}={\sqrt {\left({\frac {x-y}{2}}\right)^{2}}}={\frac {x-y}{2}}$     (where x ≥ y is assumed)

I have added the assumption in the text. Cheers. - DVdm (talk) 12:18, 22 December 2009 (UTC)

Archive 1

Archive 2

Archive 3

Archive 4

Archive 5