Question

How do you solve for x^{4}+x^{5}-1=0

Answer 1

You don't solve these kind of equations, they're too hard.

Answer 2

You can approximate x though.

Answer 3

graphically or?

Answer 4

That's one way to do it, or you ask a computer.

Answer 5

Bah too hard! :D

Answer 6

Quadradic:
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

for in the ax\(^2\) + bx + c = 0 form

Answer 7

http://www.wolframalpha.com/input/?i=solve++x^4+%2B+x^5+-+1%3D0

Answer 8

strip the coefficients off, and don't forget that if it's like this you need to pull the negatives too:

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

a = 1
b = -1
c = -1

Answer 9

what?

Answer 10

Yep you have five answers and only one of them is a real root

Answer 11

The equation has one real root and 4 imaginary roots.

Answer 12

Make a substitution if you want to use the quadratic, at first I thought that was a x^2

Answer 13

So it's messy, yes. I'd agree with you there @Thomas9

Answer 14

cool

Answer 15

You cannot use the quadratic formula here.

Answer 16

exactly quadratic cannot work...

Answer 17

Two words: Fractional exponent.

Answer 18

Agree?

Answer 19

somehow

Answer 20

http://en.wikipedia.org/wiki/Quintic_function#Solvable_quintics

Answer 21

Eliassaab i think you r so so right there...quadratic now am convinced it cant..

Answer 22

This question has been answered but not I'm going to look into this more, maybe you can't do a fractional substitution with a "quintic" polynomial... Hmm...

This is along the lines of what I was thinking
\[ax^4+bx^2+c\]
let u = x\(^2\)
\[au^2+bu+c\]

\[ax^5+bx^{2.5}+c\]
let u = x\(^{2.5}\)=\(\sqrt{x^5}\)
\[au^2+bu+c\]