• Oct 3rd 2006, 03:43 PM
sugar_babee
I've used Cardano's formula to determine an exact zero of
cube root 4 - cube root 2
The thing i need to do now is "check that it is indeed a zero of the function", I'm not sure how I'm supposed to do that. Isn't that what the point of finding the exact zero was?
If anyone could help me out here I'd really appreciate it.
• Oct 3rd 2006, 04:10 PM
topsquark
Quote:

Originally Posted by sugar_babee
I've used Cardano's formula to determine an exact zero of
cube root 4 - cube root 2
The thing i need to do now is "check that it is indeed a zero of the function", I'm not sure how I'm supposed to do that. Isn't that what the point of finding the exact zero was?
If anyone could help me out here I'd really appreciate it.

Ummm...polynomials have zeros. cubrt(4) - cubrt(2) isn't a polynomial.

-Dan
• Oct 3rd 2006, 04:14 PM
sugar_babee
The Cardano's formula is for finding one exact zero of a cubic polynomial, I'd type it out here but i don't know how to do all the curt and sqrt symbols etc, I know that this isn't wrong because it's an example question on my worksheet, I'm just not sure how to prove it like it's asking me to because I can't set the curt numbers to zero or sub them in or anything to make sure it's a root

curt 4 - curt 2 is the zero, not the polynomial, i need to know how to prove this
• Oct 3rd 2006, 04:18 PM
topsquark
Quote:

Originally Posted by sugar_babee
The Cardano's formula is for finding one exact zero of a cubic polynomial, I'd type it out here but i don't know how to do all the curt and sqrt symbols etc, I know that this isn't wrong because it's an example question on my worksheet, I'm just not sure how to prove it like it's asking me to because I can't set the curt numbers to zero or sub them in or anything to make sure it's a root

curt 4 - curt 2 is the zero, not the polynomial, i need to know how to prove this

You prove it by putting the zero back into the polynomial. As for symbols, try this:
ax^3 + bx^2 + cx + d

x = 4^(1/3) - 2^(1/3)

It'll look nasty, but that's probably the best way to type it. Give me the polynomial and I'll walk you through it.

-Dan
• Oct 3rd 2006, 04:21 PM
sugar_babee
okay, the polynomial isnt that bad, i was afraid to try putting cardano's on here :) It's f(x) = x^3 + 6x -2
• Oct 3rd 2006, 04:29 PM
topsquark
Quote:

Originally Posted by sugar_babee
okay, the polynomial isnt that bad, i was afraid to try putting cardano's on here :) It's f(x) = x^3 + 6x -2

For the first term by term I get:
[4^(1/3)-2^(1/3)]^3 = 4 - 3*4^(2/3)*2^(1/3) + 3*4^(1/3)*2^(2/3) - 2
[4^(1/3)-2^(1/3)]^3 = 2 - 3*4^(2/3)*2^(1/3) + 3*4^(1/3)*2^(2/3)

Now 4 = 2^2 so
[4^(1/3)-2^(1/3)]^3 = 2 - 3*2^(4/3)*2(1/3) + 3*2^(2/3)*2(2/3)
[4^(1/3)-2^(1/3)]^3 = 2 - 3*2^(5/3) + 3*2^(4/3)

So
x^3 + 6x - 2 =
2 - 3*2^(5/3) + 3*2^(4/3) + 6*4^(1/3) - 6*2^(1/3) - 2

= -3*2^(5/3) + 3*2^(4/3) + 6*2^(2/3) - 6*2^(1/3)

and 6 = 3*2 = 3*2^(3/3) so the above is

= -3*2^(5/3) + 3*2^(4/3) + 3*2^(5/3) - 3*2^(4/3) = 0.

-Dan
• Oct 3rd 2006, 04:42 PM
sugar_babee
Thanks a lot for your help, I understand it better now
• Oct 3rd 2006, 04:46 PM
topsquark
Quote:

Originally Posted by sugar_babee
Thanks a lot for your help, I understand it better now

I'm pleased to be of assistance. :) (I used to hate doing those myself.)

-Dan
• Oct 3rd 2006, 05:22 PM
sugar_babee
Actually, i looked it over, could you explain to me real quick what you did to get the first equation

[4^(1/3)-2^(1/3)]^3 = 4 - 3*4^(2/3)*2^(1/3) + 3*4^(1/3)*2^(2/3) - 2

I know that ^1/3 is the same as a cube root, but i don't get why you cubed the left side, and i don't see what happened to get the right side
• Oct 3rd 2006, 06:51 PM
ThePerfectHacker
Quote:

Originally Posted by sugar_babee
Actually, i looked it over, could you explain to me real quick what you did to get the first equation

[4^(1/3)-2^(1/3)]^3 = 4 - 3*4^(2/3)*2^(1/3) + 3*4^(1/3)*2^(2/3) - 2

I know that ^1/3 is the same as a cube root, but i don't get why you cubed the left side, and i don't see what happened to get the right side

He used Binomial Theorem for n=3
On a more elementary level,
(x-y)^3=x^3-3x^2y+3xy-y^3

Also
[2^(1/3)]^3=2