how to solve x^4+4x-1=0 ?

you can differentiate the expression, and then find the zero with the growth or something like that.

it depends, do you need approximative values or true values ?

i need true values

if a,b,c,d are the zeros of your polynom, then you've a+b+c+d=0, ab+bc+cd+ad = 0 , abcd = -1 and abc+abd+bcd = 4 that must be the beginning, because the values you want are1.6632 and 2.4903 ... so i think there is no easy way ...

x^(4)+4x-1=0 Use the quadratic formula to find the solutions. In this case, the values are a=1, b=4, and c=-1. x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0 Use the standard form of the equation to find a, b, and c for this quadratic. a=1, b=4, and c=-1 Substitute in the values of a=1, b=4, and c=-1. x=(-4\~((4)^(2)-4(1)(-1)))/(2(1)) Simplify the section inside the radical. x=(-4\2~(5))/(2(1)) Simplify the denominator of the quadratic formula. x=(-4\2~(5))/(2) First, solve the + portion of \. x=(-4+2~(5))/(2) Simplify the expression to solve for the + portion of the \. x=-2+~(5) Next, solve the - portion of \. x=(-4-2~(5))/(2) Simplify the expression to solve for the - portion of the \. x=-2-~(5) ~ means sqrt x~~0.24,-4.24 ~~ means approximate

it's a 4 not a 2 ??

ya ya

so your answer is not good .

Reku , the process he should understand . r u jealous with me.

? no but you can't use this formula here.

if it was x^4 + 4*x^2 -1 it would be cool, because you could use it with x^2

we can use. the quadratic formula here. hence x^4 = (x^2)^2 alright! (x^2)^2 + 4x +(-1) = 0 ax^2=1(x^2)^2 bx = 4(x) c=(-1)

i am not sure , i am a student in grade 8 . but i think i am right may be

no, because you the formula is : ax^2 + bx + c = 0 => (-b +or - sqrt(b^2 - 4*a*c)/2a Here you use it with x^2 so it would be ax^4 + bx^2 + c = 0.

hmn Reku i think now that u r right

so answer is ?

kushashwa23, i think you are wrong. It is much more difficulty question

don't talk about difficult xabord43, REKU i think here ax^2 = x^4 only if x=(x^2) let.

i assumes that it used here method Ferrari

x^4+4x-1=0 let the question be y^4+4y-1=0 ok now ax^2+bx+c= 0 ( we want the equation in the form of this ) 1(y^2)^2 + 4y - 1 = 0 we just let x be y^2 in ax^2

a - it is constant, it can't be variable

i meant for x^2 xabord try to understand!

do you know those kind of formula ? :\[(x-a)(x-b)(x-c)(x-d) = x^{4} - (a+b+c+d)x^3 + (ab+bc+cd+ad)x^2 + (abc+bcd+abd)x + abcd\]

yea reku i know

yes, Reku, i know it

ups there is a bug :s

theres 2 real and 2 complex roots x=.249 and x=-1.66

mark o. , how you did it? in Maple?

Mark O that what i am trying to tell ? they r not agree with me

I think it may solved from binom theorem

yep i said it but he want the true values :/

you people are getting it wrong use the factor theorem to solve it

find the first factor

by trial and error

not possible i think

:) the factor are -1.66325193877/...

by graphing, by quadratic formula, by maple software

it is possible ..... u pathetic people

it can not be factorized further

i m not saying to factorize it

i m saying to use factor theorem

ok

if u knw what it is then u can comment

(x+y)^4= x^4 +x^3*y + x^2 * y^2 + x* y^3 + y^4

yep, but he want the true values, so you can't graph, ese it's very simple, you can differentiate then find it with the growth of the function

\[x \approx-1.6635\]

is the final answer

lol there are other :)

there are four values of x

Yeah right

aww thnk god joe is here

http://www.wolframalpha.com/input/?i=x^4%2B4x-1 Look at how complex the roots are, there is no way anyone would be able to do this by hand unfortunately.

yes you can use newton method

thanks ! you can have the true values, but it's so awful, i made it once with ferrari, it's taking years

\[xapprox0.2490383763984\] \[xapprox0.7071067811865-1.3835510696657i\] \[xapprox0.7071067811865+1.3835510696657i\]

i am shure there is way to solve this by hand

right. there are methods of approximation we can use, but as far as being able to factor and find the exact roots, its impossible by hand.

Some sort of technology would be needed.

that are the answers

it's possible, but it's very very difficult and not very interesting

best bet is newton method thats what i used

Ya reku , u r right but if u need answers i have gave already all 4

im 100% sure its not possible by hand (to factor for exact roots). Thats just one of the things about higher degree polynomials we have to accept. We have a way to deal with quadratics for sure, we can sometimes deal with cubics, anything of higher degree is impossible unless in some special form (like difference of squares, cubes, etc.)

if you reaaallly want to do this, check ferarri method, but i won't do it ^^

yea reku ferrari method would be too long and hard

no you can't do it when the degree is 5 or more this is proven (with radicals, maybe there are other methods that we dont know yet) it s legallois theorem

http://www.proofwiki.org/wiki/Ferrari's_Method have a look t this xabord43

ups sorry abel and ruffini were the one's who found this, Le Gallois came after and found methods to show in different case, if the equation could be solve with radicals or not.

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