GoBo's Math Woes

Screw it! I'm just making a thread all about this stupid packet, which I need to have fully learned in two days!
I can't figure out this one now...

81x^3-192=0

They don't have any GCF's other then 3, which has no cubed roots. I think I understand what I'm suppose to do, normally, but this one has me beat. Help?

Well I'm terrible when it comes to math so I probably won't be able to help you too much.

By chance do you know what your doing (math terms and all that) I have my Algebra book handy so I can look it up as I know I've done that level a while ago and is no where as evil as the radicals I'm doing now.

This is on p.322-323 of our textbook. It's a problem of Difference of Cubes.

First, factor out the 3 to get:

3(27x^3 - 64)

Using this property:

a^3 - b^3 = (a - b)(a^2 +ab +b^2)

We can convert it into:

3(3x - 4)[(3x)^2 - 12x + 4^2)

3(3x - 4)(9x^2 - 12x +16) = 81x^3-192

Then you set it (3x - 4) equal to 0:

3x - 4 = 0
3x = 4
x = 4/3

For the other one, (9x^2 - 12x + 16), it cannot be factored, and you have to use the Quadratic Formula.

http://upload.wikimedia.org/math/3/e...a3bb558916.png

Substitute the values of a, b, and c for:

x = [12 + √12^2 - 4(9)(16)] / 2(9)]

x = [12 - √12^2 - 4(9)(16)] / 2(9)]

Thus:

x = (12 + 12i√3)/18
x = (12 - 12i√3)/18

oh you silly gobo xD

alright, add 192 to both sides, divide by 81. reduce to 64/27

take teh cube root of both sides. You can rewrite teh cube root of a fraction as the cube root of a numerator divided by teh cube root of a denominator, ergo root 3 64 divided by root 3 27

4/3 = x

PuPu's explanations are always so huge. I liked rubah's short and simple 8)

I just knew some math inclined people would beat me. :p

Maybe I did something wrong, but I got:

3(3x - 4)(9x² + 12x +16)

I was just told that the signs go SAME OPPOSITE PLUS, though I can't remember for the life of me why :p


For future reference:

 makes the text superscript, and  makes it subscript. It makes things much clearer in math threads n.n



EDIT:

and rubah's explanation is wrong (I think), the problem has three roots to it!


EDIT 2:

It comes down to

3=0 <--- False. Three does NOT equal zero!
x= 4/3
x= (-2 ± 2isqrt3) / (3)


...but I always make weird mistakes like adding a four instead of a six, so don't use my answers xD

5x²

...Awesome.

Quote:

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Originally Posted by Jessweeee♪

and rubah's explanation is wrong (I think), the problem has three roots to it!

---

It's not. I got the same answer. If the intended answer is positive and the variable has a odd-numbered-exponent, then the answer has to be positive. (-4/3)^3 = -81/27. I'm just going to jump to the conclusion that you are overgeneralizing the need to add multiple answers (in the form of double and triple roots), which does not apply in this situation.

If you have a graphing calculator, you can confirm this by throwing in the function, Second-Trace, and going to the Zero option (set limits as close as possible to 0 for your min, and 10 for your max in ZStandard). It will give you 1.333.

EDIT: In ZStandard, it actually looks like a line. LAWLZ

THOSE WHO EDIT FURTHER: Maybe you should consolidate all of your math troubles into one thread, Britannia-man.

Okie dokie!


EDIT:

Is it weird that I find math fun now? I am genuinely having fun doing math homework. I wish I had more, it's the only homework I actually do ;_;

Wait until you get to integrals. Mad amounts of fun to be had. Rubah can confirm.

Or LaPlace transforms. You'll be disgusted that you've been lied to throughout calculus that there's not an easier way to do everything. There is.

{Proviso: I don't actually remember how to do LaPlace transforms. But they make everything easier.}

jesse, with cubic functions, you can have either 1 real zero or 3. This is like some basic theorem of algebra or something.

for xⁿ, you can have n, n-2, n-4 etc real zeros. (you can have 0 zeros for an even function but must have at least one real zero for an odd function, that is, n is even or n is odd).

since n=3 in this case, the function is odd and can have 3 or 1 real zeros.

Quote:

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Originally Posted by Tavrobel

Wait until you get to integrals. Mad amounts of fun to be had. Rubah can confirm.

---

No. Just no.


Integrals and fun dont mix. I found them quite hard at first. Now I find them easy. But fun..?




Naow.


(with the possible exception of integration by inspection, which makes you look like a genius.)



EDIT: Is it me or was that question piss easy? It just involves a bit of rearrangement and taking the cube root. Or am I missing something/cheating?

EDITGA: Yeah, Im not cheating, and yeah its 4/3. Ahh, I love fractions. Otherwise Id be writing 1.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333 333333333333333333333...


(I think you get the point)

Quote:

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Originally Posted by Vivisteiner

Quote:

---

Originally Posted by Tavrobel

Wait until you get to integrals. Mad amounts of fun to be had. Rubah can confirm.

---

No. Just no.

EDIT: Is it me or was that question piss easy? It just involves a bit of rearrangement and taking the cube root. Or am I missing something/cheating?

EDITGA: Yeah, Im not cheating, and yeah its 4/3. Ahh, I love fractions. Otherwise Id be writing 1.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333 333333333333333333333...

---

Yes. Just yes.

Well, it's not exactly easy for someone who's never seen something like that before. It's like squares, but if you do that, then you're sure to get it wrong. Math causes psychological trauma, you know. People need examples first.

Or you could follow significant figures, and in this problem, it'd be two figures (81 has the least sig figs).