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/sci/ - Science & Math

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Anonymous Thu Feb 24 17:19:41 2011 No.2598714 [Reply] [Original]

Let's see if /sci/ really does know math.

Here's a simple problem - without looking it up, within 30 seconds of seeing this post:
Post the quadratic formula
Post regardless of whether you have the answer or not.

If you have to cheat, you should/will be expelled from /sci/ / your school.

>>	Anonymous Thu Feb 24 17:24:41 2011 No.2598732 >memorising a formula that you have no idea what it actually means You do know that it's actually the ability to derive it yourself algebraical which makes the difference between understanding and rote memory.

>>	Anonymous Thu Feb 24 17:25:41 2011 No.2598735 I have no fucking clue what a quadratic formula is. Math is for roneries.

>>	Anonymous Thu Feb 24 17:28:41 2011 No.2598755 ax^2 + bx + c = 0

>>	Anonymous Thu Feb 24 17:28:41 2011 No.2598753 (-b + (b^2 - 4ac)^(1/2))/(2a) (-b - (b^2 - 4ac)^(1/2))/(2a)

>>	Anonymous Thu Feb 24 17:30:41 2011 No.2598762 \frac{-b±\sqrt{b²-4ac}}{2a} also testing jsmath

>>	Anonymous Thu Feb 24 17:31:41 2011 No.2598771 (-b +/- sqrt (b^2 - 4ac))/2a

>>	Anonymous Thu Feb 24 17:31:41 2011 No.2598776 >>2598732 I was never taught how to derive it, just to memorize it.

>>	Anonymous Thu Feb 24 17:32:41 2011 No.2598783 <div class="math">x = -{b \over 2a} \pm \sqrt{\left({b \over 2a}\right)^2 - {c \over a}}</div>

>>	Anonymous Thu Feb 24 17:32:41 2011 No.2598784 -b+-sqrtb^2-4ac/21

>>	Anonymous Thu Feb 24 17:34:41 2011 No.2598794 All algebraic formulas may be reduced to 1=1. Problems?

>>	Anonymous Thu Feb 24 17:36:41 2011 No.2598800 >>2598762 Ok I fail, is there any tutorial on this shit anywhere?

>>	Anonymous Thu Feb 24 17:36:41 2011 No.2598803 >>2598755 >>you are banished from sci

>>	Anonymous Thu Feb 24 17:37:41 2011 No.2598808 >>2598783 wat

>>	Anonymous Thu Feb 24 17:37:41 2011 No.2598809 >>2598776 Then you don't really know math

>>	Anonymous Thu Feb 24 17:37:41 2011 No.2598814 >>2598783 HOW YOU DO THAT

>>	Josef !!nUf2NflSAyw Thu Feb 24 17:38:41 2011 No.2598819 <div class="math">x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</div> 27seconds

>>	Anonymous Thu Feb 24 17:38:41 2011 No.2598827 a^2+2ab+c^2=0

>>	Josef !!nUf2NflSAyw Thu Feb 24 17:49:41 2011 No.2598835 File: 240 KB, 4960x1754, sci_latex_tutorial.png [View same] [iqdb] [saucenao] [google] >>2598814 Somebody asked for a tutorial?

>>	Anonymous Thu Feb 24 17:50:41 2011 No.2598841 >>2598776 Then you don't really know it and should leave /sci/

>>	Anonymous Thu Feb 24 17:50:41 2011 No.2598839 x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

>>	Anonymous Thu Feb 24 17:51:41 2011 No.2598844 (-b +/- sqrt (b^2-4(a)(c)))/2a

>>	Anonymous Thu Feb 24 17:51:41 2011 No.2598847 b +- (sqrt) ba/2ab

>>	Anonymous Thu Feb 24 17:53:41 2011 No.2598850 I don't know. I stopped doing maths a long time ago and usually never post in math threads.

>>	Anonymous Thu Feb 24 17:54:41 2011 No.2598853 x = (-b +- SQRT(-4AC + B^2))/2A worked it out with paper+pen

>>	Anonymous Thu Feb 24 17:54:41 2011 No.2598855 File: 23 KB, 511x384, 1276919369481.jpg [View same] [iqdb] [saucenao] [google] >>2598755 >>2598827 /sci/ certainly does lack reading comprehension.

>>	Anonymous Thu Feb 24 17:57:41 2011 No.2598865 >>2598783 >>2598819 ok, new to /sci/ here, how do u right out equations with the math script?

>>	Anonymous Thu Feb 24 17:58:41 2011 No.2598870 I cant be bothered to remeber that shit. <span class="math">0=ax^2+bx+c=a(x+b/2a)^2+c-(b/2a)^2 \Rightarrow x=--b/2a \pm \sqrt{(b/2a)^2-c}[/spoiler]

>>	Anonymous Thu Feb 24 17:59:41 2011 No.2598876 >>2598835 <span class="math">\frac{-b±\sqrt{b^{2}-4ac}}{2a}[/spoiler] Thanks, anon.

>>	Anonymous Thu Feb 24 17:59:41 2011 No.2598883 -b +- sqroot of b^2 + 4ac all over 2a

>>	ettra transition Thu Feb 24 17:59:41 2011 No.2598878 -(p/2)±sqrt((p/2)^2-q)

>>	Anonymous Thu Feb 24 17:59:41 2011 No.2598871 File: 47 KB, 404x306, 1296014082346.jpg [View same] [iqdb] [saucenao] [google] >>2598841 >laymen aren't allowed in /sci/ >yfw

>>	Anonymous Thu Feb 24 18:01:41 2011 No.2598891 >mfw you guys haven't memorized the quintic formula

>>	Anonymous Thu Feb 24 18:02:41 2011 No.2598894 <span class="math"> (-∛((2b^3-9abc+27a^2 d+√((2b^3-9abc+27a^2 d)^2-4(b^2-3ac)^3 ))/2)-∛((2b^3-9abc+27a^2 d-√((2b^3-9abc+27a^2 d)^2-4(b^2-3ac)^3 ))/2)-b)/3a [/spoiler] cubic formula bitch

>>	Anonymous Thu Feb 24 18:03:41 2011 No.2598896 >>2598894 Now what in fucks name is that supposed to solve in the real world anyway? Just look at it.

>>	Anonymous Thu Feb 24 18:04:41 2011 No.2598902 Solving for an item's position with changing acceleration.

>>	Josef !!nUf2NflSAyw Thu Feb 24 18:06:41 2011 No.2598908 >>2598865 how do u right out equations with the math script? Ignoring the blatant typo, read my last post above.

>>	EK !EKFQOBUFnQ Thu Feb 24 18:07:41 2011 No.2598914 not cheating minus bee plus or minus the sqaure root of bee squared - 4 ay see all over 2a. amirite?

>>	Anonymous Thu Feb 24 18:08:41 2011 No.2598920 But OP, why should I memorize the quadratic formula when I can look it up in seconds?

>>	Anonymous Thu Feb 24 18:10:41 2011 No.2598930 x=[-b+/-sqrt(b^2-4ac)]/(2a)

>>	Anonymous Thu Feb 24 18:11:41 2011 No.2598933 X equals negative B, plus or minus the square root of B squared minus four A C, all over 2 A.

Anonymous Thu Feb 24 18:12:41 2011 No.2598938
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Clearly the answer is DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO DE GESO

>>	Anonymous Thu Feb 24 18:12:41 2011 No.2598940 Why not just complete the square? It's a lot more general and useful

>>	Anonymous Thu Feb 24 18:14:41 2011 No.2598946 x = (-b +/- √(b^2 - 4ac))/(2a) I hate factoring

>>	Anonymous Thu Feb 24 18:15:41 2011 No.2598951 -b = +/- Sqroot of b^2 - 4ac / 2a

>>	Anonymous Thu Feb 24 18:17:41 2011 No.2598957 >>2598835 Appreciated

>>	Anonymous Thu Feb 24 18:31:26 2011 No.2599005 wat

>>	bleedingedge !!OBNkFn9BVXZ Thu Feb 24 18:32:41 2011 No.2599011 >>2598714 x = [-b plus or minus the square root of (b^2 - 4ac)]/2a where: ax^2 + bx + c =0

>>	Anonymous Thu Feb 24 18:39:41 2011 No.2599032 x= -b + or - the square root of (b^2-4ac) all over 2a

>>	Anonymous Thu Feb 24 18:43:41 2011 No.2599049 lol i made a stupid song out of it (pop goes the wheasel) x equals negative b plus or minus the square root b square minus 4ac all over 2a~

>>	Anonymous Thu Feb 24 18:48:41 2011 No.2599068 x = (-b +/- sqrt(b^2-4ac))/2a

>>	Anonymous Thu Feb 24 19:18:17 2011 No.2599225 wat

>>	Anonymous Thu Feb 24 19:18:41 2011 No.2599228 >>2599049 full of win

>>	Anonymous Thu Feb 24 19:19:41 2011 No.2599233 >>2599049 Mr. Mac?

>>	Anonymous Thu Feb 24 19:20:41 2011 No.2599236 >>2598835 that was the shit slowing down the posting earlier today

Anonymous Thu Feb 24 19:20:41 2011 No.2599241

>read through entire thread
>expect to see actual discussion about quadratic formula, even though it's been done to death
>mfw all I see is a bunch of ego stroking 9th graders spamming the quadratic formula in butchered text trying to prove that they memorized a simple equation

ARHHGHARHGHARHHGAHRHGHARHHGHHAGHH

>>	Anonymous Thu Feb 24 19:23:41 2011 No.2599261 w/out looking at other posts: x = -b +/- [sqrt(b^2 - 4ac)/2a]

>>	Anonymous Thu Feb 24 19:25:28 2011 No.2599273 <span class="math">\int_{a}^{b} f(x)dx = F(b) - F(a)[/spoiler]

>>	Anonymous Thu Feb 24 19:29:41 2011 No.2599303 >>2599049 Which reminds me, is there no youtube account or something that makes songs out of mathematical stuff? Surely there must.

>>	anyone who doesn't know this is a fag Anonymous Thu Feb 24 19:30:41 2011 No.2599312 (-B+-sqrt((B^2)-(4ac))/(2A) +- means plus or minus

>>	Anonymous Thu Feb 24 19:31:41 2011 No.2599316 >>2598800 You just forgot the {math}{/math} script. With brackets, though. It'll look like this: <span class="math">\frac{-b±\sqrt{b²-4ac}}{2a}[/spoiler]

>>	Anonymous Thu Feb 24 19:33:41 2011 No.2599322 i made a song so I could remember it, but I haven't used it in so long so I wonder if it's right x equals -b, plus or minus the square root of b squared minus 4ac all over 2 a sung in the tune of pop goes the weasel

>>	Anonymous Thu Feb 24 19:40:41 2011 No.2599347 y=mx^2+bx+c

Anonymous Thu Feb 24 19:53:41 2011 No.2599406

I remember back in highschool

I learnt this, and since I was unattentive in class I talked about it to a friend, and he said there was a simple explanation/way to figure it out without looking it up.

Untill this day I didn't do any research about it. Anyone care to explain where this formula comes from?

>>	Anonymous Thu Feb 24 20:00:41 2011 No.2599439 <span class="math"> \mathrm{testing} [/spoiler]

>>	Anonymous Thu Feb 24 20:06:41 2011 No.2599475 >>2599233 does Mr. Mac have a thin beard, all black hair, and did he teach in Nova Scotia?

>>	Anonymous Thu Feb 24 20:10:41 2011 No.2599494 >>2599406 Complete the square

>>	Anonymous Thu Feb 24 20:13:41 2011 No.2599507 -b +/- ((b^2 - 4ac)^(1/2))/2a Terrible notation but I think that's at least close to correct.

>>	Crimean Emperor !RZND91lf7s Thu Feb 24 20:15:41 2011 No.2599522 -b(plus or minus)(sqrt)(b^2-4ac)/2a

Anonymous Thu Feb 24 20:17:41 2011 No.2599534

>>2599507

Terrible notation indeed, especially as it implies that only the part of the equation in brackets is being divided by 2a.

I'm starting to learn C++ and I used this when I was writing a program to solve quadratic equations:

root1 = (-B + sqrt(B*B-4*A*C))/(2*A);
root2 = (-B - sqrt(B*B-4*A*C))/(2*A);

>>	Anonymous Thu Feb 24 20:20:41 2011 No.2599550 x equals negative b plus or minus the square root of b squared minus four ac, all over 2a

>>	Anonymous Thu Feb 24 20:22:41 2011 No.2599562 File: 351 KB, 1599x1200, 1269997955782.jpg [View same] [iqdb] [saucenao] [google] >>2599534 Your face when roots turn out to be complex or a is very small

>>	Anonymous Thu Feb 24 20:22:41 2011 No.2599566 >>2598783 a challenger appears

Anonymous Thu Feb 24 20:30:41 2011 No.2599615

>>2599562

I only wrote it to solve the equation when there is at least one real root (it prints: "This equation has no real roots" when (B*B < 4*A*C)). It seems to work with values like 0.01 for A, although it messes up with extremely large or small numbers

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