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/sci/ - Science & Math

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3844305 No.3844305 [Reply] [Original]

how the fuck do i solve:
x = x^3 + 1

i tried it but i just keep going in circles no matter what i do.

>> No.3844309

i would have done so much better with my higher math had i 4chan's backing

>> No.3844312

>>3844305
subtract the x on the left, factor, solve for 0

>> No.3844321

x=1

>> No.3844315

DEYR BOTH THE SAME VARIABLES

>> No.3844322

1 = x - x^3

>> No.3844323
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3844323

>>3844321
>mfw 1=2

>> No.3844328

(x-1)(x+1)=(1/x)

>> No.3844329
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3844329

>>3844323
shit, just thought that over

>> No.3844330

>>3844312
(x+1.32472) (x-(0.662359+0.56228 i)) (x-(0.662359-0.56228 i))

>> No.3844335
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3844335

-1.32

>> No.3844338

inb4 "i'm so smart faggots" spoonfeed OP the answer.

>> No.3844347
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3844347

>>3844330
>>3844335

>mfw you use decimals for algebra

>> No.3844355

>>3844347
-(2/(3 (9-sqrt(69))))^(1/3)-(1/2 (9-sqrt(69)))^(1/3)/3^(2/3)

come at me bro

>> No.3844354

>>3844305
wolframalpha.com

>> No.3844373 [DELETED] 

>>3844305
the solution is somewhere between -1 and 0 so try newtons method


x-(x^3-x+1)/(3x^2-1)

>> No.3844370
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3844370

>>3844355
how the hell did you manage that? This question was a bonus on my calc 1 exam today and i clearly didn't get it

>> No.3844377
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3844377

your options

a) Guess the answer
b) know a lot about modular spaces and Galois theory and feel the answer
c) use the general formula
http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function
which exists for polynomials up to 4th oder
d) use da computa
http://www.wolframalpha.com/input/?i=Solve%5Bx%3D%3Dx%5E3%2B1%2Cx%5D

>> No.3844380

>>3844373
i meant the solution is between -2 and -1

>> No.3844389

the solution is somewhere between -2 and -1 so try newtons method


x-(x^3-x+1)/(3x^2-1)

>> No.3844390

>>3844377
>http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function
how the fuck is someone suppose to remember that shit

>> No.3844399
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3844399

>>3844390
I wrote it down so I don't have to remember.

>> No.3844405

>>3844399
what about when you can't use your reference sheet? where's your god now?

>> No.3844406

NEWTONS METHOD YOU AUSTITIC FUCKS

x-f(x)/f'(x)

>> No.3844424

>>3844406
kid's in calc 1 so probably hasn't yet done derivatives. you're an idiot

>> No.3844432

>>3844405
>mfw you probably learned something in your class which could help you with this but you were too lazy to pay attention.

>> No.3844429

>>3844305

>x - 1 = x ^ 3

I don't think so Tim.

>> No.3844451

>>3844429
dur water neg numbahs

>> No.3844489

>mfw niggers shout newtons method when you are supposed to find exact values

>> No.3844493

x-1=x^3
(x-1)^(1/3)=x
((x-1)^(1/3)-1)^(1/3)=x
(((x-1)^(1/3)-1)^(1/3)-1)^(1/3)=x
((((x-1)^(1/3)-1)^(1/3)-1)^(1/3)-1)^(1/3)=x
Repeat
???
PROFIT!!!

>> No.3844501

>>3844390
program it into your graphing calculator?

>> No.3844508

>>3844501
no calculators allowed

>> No.3844514

It has no real roots, use the cubic formula

>> No.3844519

>>3844508
then fucking memorize it then, faggot
you know quadratic formula, man the fuck up and memorize cubic formula

>> No.3844523
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3844523

>>3844519

>> No.3844543

Vieta's formula

>> No.3844553

>>3844489
>implying you learn the cubic formula in calculus

itt:high school kiddies who have yet to take or pass calculus 1


newtonds method would have been fine

>> No.3844560

>>3844553
> newtond's

Leave.

>> No.3844563

>>3844560
what? you mad that you don't know about newton's method? I bet you go to CC faggot

>> No.3844581

x = x^3 + 1
x(x) = x(x^3) + 1
x^2 = x^4 + 1
x^4 - x^2 + 1 = 0

You can do the rest from there, should be much more clearer.

Start from -
>x^4 - x^2 + 1 = 0

>> No.3844592
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3844592

>>3844581

> mfw x is distributed to half of the right side

>> No.3844595

http://www.wolframalpha.com/input/?i=x+%3D+x^3+%2B+1

>> No.3844597

>>3844592
Oh shit.
*x^4+x^2+x = 0

It's 2am.

>> No.3844598

>>3844581
x = x^3 + 1
x(x) = x(x^3) + 1
x^2 = x^4 + 1
x^4 - x^2 + 1 = 0
x^2(x^2-1) +1 = 0
x^2 (x+1)(x-1) = -1

>> No.3844620

x-1=x^3
(x-1)^(1/3)=x
((x-1)^(1/3)-1)^(1/3)=x
(((x-1)^(1/3)-1)^(1/3)-1)^(1/3)=x
((((x-1)^(1/3)-1)^(1/3)-1)^(1/3)-1)^(1/3)=x
Plug a rough approximate value in for x since the error should be minimized., eg. x=0
x approximately = ((((0-1)^(1/3)-1)^(1/3)-1)^(1/3)-1)^(1/3)=-((((1)^(1/3)+1)^(1/3)+1)^(1/3)+1)^(1/3)=-1.322

actual value is -1.324

>> No.3844623

>>3844581
>>3844597

How does that even help with this question

>> No.3844624

Use Cardano.
<div class="math">x^3-x+1=0</div>
Substitute <span class="math">x=w+\frac{1}{3w}[/spoiler]:
<div class="math">w^3 + \frac{1}{27w^3} + 1 = 0</div>
Multiply both sides by <span class="math">w^3[/spoiler] to reduce this to a quadratic in <span class="math">w^3[/spoiler].

>> No.3844632

>>3844620
what's that called?

>> No.3844633

x^3-x=-1
x(x^2-1)=-1
x1=-1
x2,3=0

>> No.3844637

>>3844633
x2,3=0 implies 0=1 dumb shit

>> No.3844676

>>3844632
No clue, I came up with it just now.