Question

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

Answer 1

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

Answer 2

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

Answer 3

i need true values

Answer 4

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 ...

Answer 5

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

Answer 6

it's a 4 not a 2 ??

Answer 7

ya ya

Answer 8

so your answer is not good .

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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)

Answer 13

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

Answer 14

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.

Answer 15

hmn Reku i think now that u r right

Answer 16

so answer is ?

Answer 17

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

Answer 18

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

Answer 19

i assumes that it used here method Ferrari

Answer 20

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

Answer 21

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

Answer 22

i meant for x^2 xabord try to understand!

Answer 23

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

Answer 24

yea reku i know

Answer 25

yes, Reku, i know it

Answer 26

ups there is a bug :s

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

I think it may solved from binom theorem

Answer 31

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

Answer 32

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

Answer 33

find the first factor

Answer 34

by trial and error

Answer 35

not possible i think

Answer 36

:) the factor are -1.66325193877/...

Answer 37

by graphing, by quadratic formula, by maple software

Answer 38

it is possible ..... u pathetic people

Answer 39

it can not be factorized further

Answer 40

i m not saying to factorize it

Answer 41

i m saying to use factor theorem

Answer 42

ok

Answer 43

if u knw what it is then u can comment

Answer 44

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

Answer 45

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

Answer 46

\[x \approx-1.6635\]

Answer 47

is the final answer

Answer 48

lol
there are other :)

Answer 49

there are four values of x

Answer 50

Yeah right

Answer 51

aww thnk god joe is here

Answer 52

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.

Answer 53

yes you can use newton method

Answer 54

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

Answer 55

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

Answer 56

i am shure there is way to solve this by hand

Answer 57

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.

Answer 58

Some sort of technology would be needed.

Answer 59

that are the answers

Answer 60

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

Answer 61

best bet is newton method thats what i used

Answer 62

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

Answer 63

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.)

Answer 64

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

Answer 65

yea reku ferrari method would be too long and hard

Answer 66

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

Answer 67

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

Answer 68

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.