equation number 2

RauLiTo · 2006-02-06 08:59:50

try to get X :

[ X - 1 / X ] ³ + 3 [ X - 1 / X ] + 30 = 0
X = ?

it look easy but i don't know whats wrong with this mind

mathsyperson · 2006-02-06 10:37:43

I think the best way of doing this would be to solve y³ + 3y + 30 = 0, then use that answer to solve y = x-1/x.

Unfortunately, I have no idea how to solve cubic equations. But hopefully I've put you on the right track.

irspow · 2006-02-06 11:29:37

Here's my stab.

I took your equation and expanded it to;

(x^6 + 30x - 1) / x^3 = 0 ( I don't feel like typing out the steps, I'm sure you can do it too.)

So this only equals zero if;

x^6 + 30x - 1 = 0

Let x^3 = y, then;

y^2 + 30y - 1 = 0

Using the quadratic equation gives;

y = .033296378.. and -30.03329638...

Since y = x^3, x = y^1/3

x = .321710818... and -3.1081632...

edit*

I checked this with my TI-89 and my answer was confirmed correct. The slight error from being exact is only from rounding the value of x.

The precise answer is;

-(√(226) + 15)^(1/3) and (√(226) - 15)^(1/3)

Which agrees with what I had earlier.

We don' need no stinkin' calculators....

Last edited by irspow (2006-02-06 13:05:40)

krassi_holmz · 2006-02-06 19:08:01

Right.
Actually there are several more complex roots:

Last edited by krassi_holmz (2006-02-06 19:14:31)

kempos · 2006-02-06 21:03:13

i don't get this step:

x^6 + 30x - 1 = 0

Let x^3 = y, then;

y^2 + 30y - 1 = 0

why do you have 30y if y = x^3?

RauLiTo · 2006-02-06 21:13:02

thanks everybody for helping as kempos said i dind't get that one

RauLiTo · 2006-02-06 21:14:10

krassi_holmz ... i want the way you get the answer please man

krassi_holmz · 2006-02-06 21:41:17

I used Mathematica.
Must write something like this:
Solve[( X - 1 / X ) ³ + 3 ( X - 1 / X ) + 30 == 0,X]

irspow · 2006-02-07 08:23:00

Sorry guys, it was a typing mistake. The "30x" should have been 30x^3. I didn't even noticed that I typed it that way. If I actually used the 30x, I would have never have gotten the answer.
Sorry, again.

Math Is Fun Forum

#1 2006-02-06 08:59:50

equation number 2

#2 2006-02-06 10:37:43

Re: equation number 2

#3 2006-02-06 11:29:37

Re: equation number 2

#4 2006-02-06 19:08:01

Re: equation number 2

#5 2006-02-06 21:03:13

Re: equation number 2

#6 2006-02-06 21:13:02

Re: equation number 2

#7 2006-02-06 21:14:10

Re: equation number 2

#8 2006-02-06 21:41:17

Re: equation number 2

#9 2006-02-07 08:23:00

Re: equation number 2

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