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Topic review (newest first)

irspow
2006-02-08 07:23:00

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.

krassi_holmz
2006-02-07 20:41:17

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

RauLiTo
2006-02-07 20:14:10

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

RauLiTo
2006-02-07 20:13:02

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

kempos
2006-02-07 20:03:13

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?

krassi_holmz
2006-02-07 18:08:01

Right.
Actually there are several more complex roots:






irspow
2006-02-07 10:29:37

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....

mathsyperson
2006-02-07 09:37:43

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.

RauLiTo
2006-02-07 07:59:50

try to get X smile :

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

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

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