Math Help, Please

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Spleen
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Math Help, Please

Unread postby Spleen » Sat Aug 23, 2003 12:56 pm

The summer assignment for people going into geometry is a sheet of algebra problems. A friend of mine and I found one particular problem that neither of us could figure out using what we learned in Algebra I. The problem is a trinomial, and you must solve it for x.

The problem is:
12x^2 - 6x + 21 = 0

The problem we're having is probably that we've had months to let all math knowledge cultivated in eighth grade to rot. Hopefully someone here has not had likewise happen to them and can help with this problem. <p>_________________
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Lord McBastard
 

Re: Math Help, Please

Unread postby Lord McBastard » Sat Aug 23, 2003 1:41 pm

x = 4 <p><div style="text-align:center">
Yeah with enough cheese we could blow up just about anything.-Tyler Durden (misquoted)</div></p>

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Re: Math Help, Please

Unread postby Spleen » Sat Aug 23, 2003 2:22 pm

How'd you get that? What are you supposed to put in the parenthesis when you factor the trinomial? <p>_________________
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Re: Math Help, Please

Unread postby Idran1701 » Sat Aug 23, 2003 2:59 pm

Did you learn the quadratic formula in your last math class? If you did, try putting the coefficients into it and see what you get. At the very least, check the discriminant. And ignore LMB's answer. <p>

"Never let your morals get in the way of doing what is right" - Salvor Hardin</p>

Archmage144
 

Re: Math Help, Please

Unread postby Archmage144 » Sat Aug 23, 2003 3:35 pm

Idran is a wise man. <p>
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Re: Math Help, Please

Unread postby Spleen » Sat Aug 23, 2003 4:59 pm

I forgot the quadratic formula, and I don't know what a discriminant is. <p>_________________
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Re: Math Help, Please

Unread postby Idran1701 » Sat Aug 23, 2003 5:22 pm

Quadratic formula:

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

The discriminant is b^2 - 4ac. If it's positive, you have two answers. Zero, and you have one. Negative, and you have no answers in the real number system. <p>

"Never let your morals get in the way of doing what is right" - Salvor Hardin</p>

Phil
 

Dee Piddely Bo skiddely.

Unread postby Phil » Sat Aug 23, 2003 9:10 pm

Er, I'm thinking maybe Spleen needs some really basic instructions, seeing as how he hasn't had freshman algebra. I apologize in advance if this is too basic, I'm writing this assuming you know nothing at all, because that's better than assuming you know too much and not explaining everything as much as you need.

Now, before I start, I should let you know that your particular equation cannot be solved by factoring, because the solutions are not real numbers - Idran has already explained why using the quadratic formula. The left and right numbers are both very large, so the number under the square root sign is negative. The b^2 - 4ac is really the only number you need to find if you only want to know if the solutions are real.

12x^2 - 6x + 21 = 0
This will give you 36 - 4(12)(21) in the quadratic formula, which is clearly negative, which means your problem cannot be solved by factoring.


But I'll give you some factoring tips with a new example.

x^2 - x - 20 = 0

Step 0: Make sure the thing equals zero. If it doesn't, you'll have to move things to the other side of the equals sign until you have one side empty.

1. Look at the number in front of the x^2 and the number with no x term in it. These will give you guidelines for what you can possibly put where.

2. You know that you're going to have to break this thing into something that looks like (0x + 0) (0x + 0), where of course the 0's are other numbers. Now, it's the x^2 and zero-x term that are going to give you your best clues. Here's what you can know from them.

a. The numbers you're going to put in the left part of the parenthesis (The ones with X's next to them) must somehow multiply to 1 (the number on the furthest left). This pretty much means you don't have to stick anything on the left side of your solution.


b. The numbers you're going to put in the right side of the parenthesis (with no X's) must somehow multiply to -20.
Possible combinations are
1 and -20 or -1 and 20
2 and -10 or -2 and 10
4 and -5 or -4 and 5

3. Okay, so you know what can be on the left and right sides - so what? I mean, there can be only one answer and I've given you a lot of possible ones. Well, you see, there's one catch, and it's going to make this problem much harder. What I'm talking about is the middle number (with one X) that I strategically avoided mentioning until now.

The third and hardest condition you have to meet is this: The product of the outer numbers plus the product of the inner numbers has to equal negative 1, because you have that -x in the middle. This is going to require of trial and error usually.

4. So since you know the total you're looking for is -1, select two numbers that can go on the right side that multiply to -20 and add to negative 1. If you look at the choices above, the pair of [4 and -5] will do quite nicely.

4. In general, here's some other stuff to remember:

a. Make sure to watch the signs on all numbers. The first number will always be positive (in most math books) but the others may not be. If you try to work it with the incorrect signs, you'll get dead-ended every time.

b. And make sure to remember that two negatives multiplied together is a positive - if the zero-x term in the problem is positive, then that could mean either two positives OR two negatives on the right side of both parenthesis in your solution.

I hope I haven't spent the last half hour telling you what you already know.


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Re: Dee Piddely Bo skiddely.

Unread postby Spleen » Sat Aug 23, 2003 11:02 pm

Phil, factoring is all I do know, and that's my problem. Like you said, it doesn't work in this problem. I took Algebra I after all, I'm not uniformed. But thanks for the attempt anyway, man. <p>_________________
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-Aku, Samurai Jack</p>Edited by: [url=http://pub30.ezboard.com/brpgww60462.showUserPublicProfile?gid=spleeninfinity13>SpleenInfinity13</A]&nbsp; Image at: 8/23/03 11:06 pm

Rhelian Alto
 

Re: Dee Piddely Bo skiddely.

Unread postby Rhelian Alto » Sun Aug 24, 2003 2:25 am

Like Idran said before, it's just the quadratic formula. We had this DRILLED into us by our eighth grade algebra teacher, and it's come in handy.

Here's the low down:

You have this.
12x^2 - 6x + 21 = 0

The quadratic formula is:
x = (-b +/- sqrt(b^2 - 4ac)) / 2a

Now, here's where we plug numbers in. Your major variables will be a, b, and c. Using these, we get (usually) two values for x. Here's how to get find the variables.

A is the number that you multiply x^2 by. In this case, 12x^2, it's 12. Often you'll find that there is no multiplier, in which case, you have 1, because 1x^2 is the same as x^2.

B is the number that you multiply your second x by. So, -6x makes this number -6.

C is the last number in the equation. +21.

So now we have a = 12, b = -6, and c = 21. Let's plug these in, shall we?

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

x = (-[-6] +/- sqrt([-6]^2 - 4[12][21])) / 2[12]

x = (6 +/- sqrt(36 - 1008)) / 24

x = (6 +/- sqrt(-972)) / 24

Notice the sqrt(-972)? We can't take a square root of a negative number, so the problem is unsolvable in the real number system. We can use i to represent non-real numbers, but that's a whole new can of worms. Suffice it to say, this problem's answer cannot be expressed in real numbers. As Idran also said, the discriminant (b^2 - 4ac) will be positive to get two answers, zero for one answer, or negative for a non-real answer. In this case, our discriminant is negative.

This ends my lesson on the quadratic formula, and I hope my elaboration helped you. You can use this formula on ANY equation that follows the pattern of x^2 + x + 1 = 0 or something to that effect. With more experience you'll be able to pick these out and work with them, no problem. I hope I didn't just confuse the hell out of you or say something wrong.

EDIT: I also hope I'm not acting condescending towards you - that's not my intention. It just seemed like you needed elaboration on Idran's advice.

EDIT Mk II: The asnwers are (hopefully):
(6 + 31.177i) / 24 and (6 - 31.177i) /24

Please, someone call me on this stuff if I'm wrong. It behooves me to know it as well. <p>

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A waist is a terrible thing to mind.
Eat well, stay fit, die anyway.</p>Edited by: [url=http://pub30.ezboard.com/brpgww60462.showUserPublicProfile?gid=rhelianalto>Rhelian]&nbsp; Image at: 8/24/03 2:32 am

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Spleen
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Re: Dee Piddely Bo skiddely.

Unread postby Spleen » Mon Aug 25, 2003 10:47 pm

So the answer doesn't exist. No fucking wonder we couldn't get the answer. Thanks for the help, everybody. <p>_________________
FEEL MY WRATH!
...it's very fuzzy.

"Samurai, samurai, why won't you DIE?!"
-Aku, Samurai Jack</p>

Lord McBastard
 

Re: Dee Piddely Bo skiddely.

Unread postby Lord McBastard » Mon Aug 25, 2003 11:26 pm

I'm telling you.

X = 4.

Jesus! My math is alot simpler than your fancy formulas. <p><div style="text-align:center">
Yeah with enough cheese we could blow up just about anything.-Tyler Durden (misquoted)</div></p>

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Re: Dee Piddely Bo skiddely.

Unread postby SorataYuy » Tue Aug 26, 2003 12:46 am

*swirly-eyed now, after trying to read all that, falls over on his face* ooooooo... oi... how the hell do you guys keep all that stuff straight? I'd get a headache if I could look at it without my self-preservation instinct kicking in and my brain turning off. <p>I BUKKAKE FOR JUSTICE!!!</p>

Wolfbelly
 

Re: Dee Piddely Bo skiddely.

Unread postby Wolfbelly » Tue Aug 26, 2003 4:40 am

Sweet mother of crap ... I used to know how to do these without batting an eye ... and now, now ... I get lost at the word 'trinomial!'

*Breaks down sobbing*


Archmage144
 

Re: Dee Piddely Bo skiddely.

Unread postby Archmage144 » Tue Aug 26, 2003 10:18 am

Imaginary numbers are fun!

*plots the answers to the problem on the imaginary number line* <p>
<div style="text-align:center">Image</div>

</p>

Kakita Ki
 

Re: Dee Piddely Bo skiddely.

Unread postby Kakita Ki » Tue Aug 26, 2003 12:39 pm

Oh geez, I don`t remember any of this and I have geometry in a week. <p>
Sig coming kinda soon...<span style="font-size:xx-small;">never</span>
</p>

Lord McBastard
 

Re: Math Help, Please

Unread postby Lord McBastard » Tue Aug 26, 2003 1:53 pm

The answer: I don't care. Would have also been acceptable. <p><div style="text-align:center">
Yeah with enough cheese we could blow up just about anything.-Tyler Durden (misquoted)</div></p>

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Re: Math Help, Please

Unread postby Banjooie » Sun Aug 31, 2003 1:55 am

Canadian schools suck or something

Grade 10 math didn't even touch quadratic formula. <p><Chat> <Matto says, "What's up?"
<Chat> <Prince_Herb says, "Angst."
<Chat> <Prince_Herb says, "Drama."
<Chat> <Prince_Herb says, "Betrayal."
<Chat> <Prince_Herb says, "Plushies."</p>


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