Eyes on Final Fantasy Forums > Reference > Study Hall > GoBo's Math Woes

[Goldenboko]

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?

[Bahamut2000X]

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.

[PuPu]

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.



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

[RubyLapiz]

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

[Goldenboko]

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

[Bahamut2000X]

I just knew some math inclined people would beat me.

[Jessweeee♪]

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


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

[Goldenboko]

5x²

...Awesome.