Hi all! How do I get the x-intercept of -x^3 +3x^2 +2 ? I understand you set it equal to zero, but after that I'm stuck :/

Question

anonymous · Answer

so basically u got stuck while solving the cubic equation

anonymous · Answer

yes

anonymous · Answer

So I have -x^3 +3x^2 =-2

anonymous · Answer

then factor?

anonymous · Answer

No, because it's not equal to 0

anonymous · Answer

hmm

ganeshie8 · Answer

-x^3 + 3x^2 = -2

$x^3 - 3x^2 -2 = 0$

eliminate x^2 term, 
put y = x-1  => $x = y+1$

ganeshie8 · Answer

this is going to be a long process, u may give up if u dont wanto deal wid all the mess + complex numbers that wil come on the way..  :)

ganeshie8 · Answer

u can simply approximate the root wid newton if u want

anonymous · Answer

Hmm, I don't know this method

anonymous · Answer

how do I do it?

ganeshie8 · Answer

pick one below :-
1) newton approximation (calculus)
2) algebraic solution

anonymous · Answer

1

ganeshie8 · Answer

very wise :)

anonymous · Answer

How do I start?

anonymous · Answer

like this? http://www.youtube.com/watch?v=1uN8cBGVpfs

ganeshie8 · Answer

yes, lets start with x = 3 maybe

anonymous · Answer

yeah, I was thinking between 3.1 and 3.2, is that good?

ganeshie8 · Answer

$f(x) :   x^3 - 3x^2 -2 $

anonymous · Answer

ah ok

ganeshie8 · Answer

$\large  x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

ganeshie8 · Answer

start wid $x_0 = 3$

anonymous · Answer

ok

ganeshie8 · Answer

its a recursive formula, the approximation gets better as the iterations increase

anonymous · Answer

mhm

anonymous · Answer

attachment

anonymous · Answer

is that the algebraic solution?

anonymous · Answer

i used wolfram coz it was beyond my level .!

ganeshie8 · Answer

yep, u dont want to do that.. not in ur scary night mares :|

anonymous · Answer

XD it's scaaary

anonymous · Answer

so, the newton method will yield the same/a similar result?

ganeshie8 · Answer

lets see :)

$\large  x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

$f(x) :   x^3 - 3x^2 -2 $
$f'(x) :   3x^2 - 6x$

Iteration 1 :
$\large  x_1 = 3 - \frac{f(3)}{f'(3)}$

ganeshie8 · Answer

thats the approximation after first iteration

anonymous · Answer

I don't understand this part of the notation which says frac

ganeshie8 · Answer

oh you're not seeing latex, there is a problem with latex in this site for some users

anonymous · Answer

ah, what must I do?

ganeshie8 · Answer

come here http://www.twiddla.com/1462458

anonymous · Answer

k

anonymous · Answer

thanks again everyone for your help!

ganeshie8 · Answer

u wlc :)