Question

Don't troll me, I know I suck at math.
ATTEMPT NUMBER 6:

\(\huge\color{blue}{\huge {\bbox[5pt,cyan,border:2px solid purple]{f(x)=x^3+x+1}}}\)

Answer 1

\(\huge\color{blue}{\huge {\bbox[5pt,cyan,border:2px solid purple]{PLEASE~~~~EXPLAIN}}}\)
\(\huge\color{blue}{\huge {\bbox[5pt,cyan,border:2px solid purple]{How~~do~~I~~find~~the~~roots?}}}\)

Answer 2

kk

Answer 3

sorry I suck at math also

Answer 4

Sorry, that's too advanced for me and I suck at math too.

Answer 5

no idea

Answer 6

You find roots by setting it equal to zero. This is tricky to factor, since it is x^3. It actually doesn't factor. But if you graph it and find the x-intercept (the zero/root), you will find x=-.68233

Answer 7

So I can only approximately find the roots, I can never actually find the roots for sure, like an exact value (sqrt5 -4i or integer or something like that)?
It will depend on how accurately will I graph it?

Answer 8

can u use approximation ??

Answer 9

there is no intrger sol for sure but u can a proximate..

Answer 10

approximation? I was never good at mathematical terms, what do you mean, perhaps I might know.
Also, is there an exact answer to the problem?

Answer 11

May it be -5i or sqrt4+3i or anything, but exact value.

Answer 12

it might have an exact sol but not integer

Answer 13

yeb it could be complex or irrational

Answer 14

Whatever the exact value is, how do you find it?

Answer 15

ok do u know kardano method ??

Answer 16

No, but I think ganeshie8 showed to me, I don't remember the formula, and I've never done it. It would be pretty difficult to teach me, being that I know nearly nothing.

Answer 17

its for
x^3=px+q
\[x=\sqrt[3]{\frac{ q }{ 2 }+\sqrt{\frac{ q^2 }{ 4 }-\frac{ p^3 }{ 27 }}}+x=\sqrt[3]{\frac{ q }{ 2 }-\sqrt{\frac{ q^2 }{ 4 }-\frac{ p^3 }{ 27 }}}\]

Answer 18

ok gust apply the formula and ul get a sol :)

Answer 19

oh i type it rongly sry ..

Answer 20

i need time to calculate it....

Answer 21

rongly?

Answer 22

\[x=\sqrt[3]{\frac{ q }{ 2 }+\sqrt{\frac{ q^2 }{ 4 }-\frac{ p^3 }{ 27 }}}+\sqrt[3]{\frac{ q }{ 2 }-\sqrt{\frac{ q^2 }{ 4 }-\frac{ p^3 }{ 27 }}}\]

Answer 23

oh, w/o x=

Answer 24

take ur time its oly calculating =)

Answer 25

p=-1
q=-1

Answer 26

What? I though it's +1 and +1

Answer 27

x^3=px+q
ur equation is
x^3+x+1
so
x^3=-x-1
q=-1
p=-1

Answer 28

I'll be back, 2 minutes...

Answer 29

A plot and solution using Mathematica 8 Home Edition is attached.

Answer 30

Sorry, there are two mis-spellings in solomonzelman.pdf

Answer 31

Lol u say u suck at math when i think ur like a math teacher...

Answer 32

SO to find a root of this function all you're trying to do is find where the graph hits zero

Answer 33

How does one write in such cool latex O_O

Answer 34

thank you, my latex aren't cool.....
anyway,
http://openstudy.com/study#/updates/51fbcbade4b0cc46c14a461d

Answer 35

I just combined them.

Answer 36

\(\huge\color{magenta}{\huge {\bbox[5pt,yellow ,border:2px solid blue ]{~•~~‿~•~}}}\)

Answer 37

http://www.wolframalpha.com/input/?i=x%5E3%2Bx%2B1%3D0

you can click exact forms. They are very ugly.

Answer 38

it's +31, not just 1.

Answer 39

I see, but how does he get it?

Answer 40

What does shamil 98 mean by writing in cool latex?

Answer 41

Idk, he thought my latex were cool perhaps, but they're not.

Answer 42

"it's +31, not just 1."

no it isn't, in your first post.

Answer 43

\(\Huge{\color{purple}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{cyan}{\bigstar}\color{purple}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{cyan}{\bigstar}\color{purple}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\\\color{white}{.}\\\Huge\sf\color{magenta }{~~f(x)=x^3+x+31~\\}\\\color{white}{.}\\\\\Huge{\color{purple}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{cyan}{\bigstar}\color{purple}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{cyan}{\bigstar}\color{purple}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\)

Answer 44

http://www.wolframalpha.com/input/?i=x%5E3%2Bx%2B31%3D0

Answer 45

I re-inputted (if that makes sense) and I see the answer, but I need to know how to do it.

Answer 46

I am going to play basketball right now, thank for trying to help everyone!
It's waiting for me!!!!!!!!!!!!!!
\(\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\\\color{white}{.}\\\Huge\sf\color{blue}{~~~~Have~~~a~~~good~~~night!~\\}\\\color{white}{.}\\\\\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\)

Answer 47

lol follow the steps http://www5a.wolframalpha.com/Calculate/MSP/MSP6131f9eb899264a439e00003fhe00c8f7ceg1ed?MSPStoreType=image/png&s=9&w=500&h=3867

Answer 48

if you look at the steps for exact solution in wolfram it breaks down how to do it
it uses a variable transformation then back substitution

Answer 49

normally when no rational zero exists , approximation from calculator suffice or if you know calculus , Newtons method is great for finding irrational zeros

Answer 50

Yeah, I kind of got it.