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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i need true values

OpenStudy (anonymous):

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 ...

OpenStudy (anonymous):

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

OpenStudy (anonymous):

it's a 4 not a 2 ??

OpenStudy (anonymous):

ya ya

OpenStudy (anonymous):

so your answer is not good .

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

hmn Reku i think now that u r right

OpenStudy (anonymous):

so answer is ?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i assumes that it used here method Ferrari

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i meant for x^2 xabord try to understand!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yea reku i know

OpenStudy (anonymous):

yes, Reku, i know it

OpenStudy (anonymous):

ups there is a bug :s

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I think it may solved from binom theorem

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

find the first factor

OpenStudy (anonymous):

by trial and error

OpenStudy (anonymous):

not possible i think

OpenStudy (anonymous):

:) the factor are -1.66325193877/...

OpenStudy (anonymous):

by graphing, by quadratic formula, by maple software

OpenStudy (anonymous):

it is possible ..... u pathetic people

OpenStudy (anonymous):

it can not be factorized further

OpenStudy (anonymous):

i m not saying to factorize it

OpenStudy (anonymous):

i m saying to use factor theorem

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

if u knw what it is then u can comment

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

\[x \approx-1.6635\]

OpenStudy (anonymous):

is the final answer

OpenStudy (anonymous):

lol there are other :)

OpenStudy (anonymous):

there are four values of x

OpenStudy (anonymous):

Yeah right

OpenStudy (anonymous):

aww thnk god joe is here

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

yes you can use newton method

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i am shure there is way to solve this by hand

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

Some sort of technology would be needed.

OpenStudy (anonymous):

that are the answers

OpenStudy (anonymous):

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

OpenStudy (anonymous):

best bet is newton method thats what i used

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yea reku ferrari method would be too long and hard

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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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