Question

`x^4-4x^3+x^2+4x+1=0`. How to solve for x?

Answer 1

`x=(1+-sqrt5)/2,x=(3+-sqrt13)/2`

Explanation:

Since this quartic does not have rational roots (and I cannot be bothered with the formulas), we start by using Newton's method to approximate the roots:

`x~~-0.303`

`x~~-0.618`

`x~~1.618`

`x~~3.303`

Of these, we find that `x~~-0.618` and `x~~1.618` stand out. We recognize these as the golden ratio:
`x=(1+-sqrt5)/2`

We can also verify that they are roots by plugging them into the equation, but you can just take my word that they are indeed roots.

This means that the following is a factor of the equation:
`(x-(1+sqrt5)/2)(x-(1-sqrt5)/2)=`

`=((x-1/2)+sqrt5/2)((x-1/2)-sqrt5/2)=`

`=(x-1/2)^2-(sqrt5/2)^2=x^2-x+1/4-5/4=`

`=x^2-x-1`

Since, we know `x^2-x-1` is a factor, we can use polynomial long division to find out the remainder and rewrite the equation like so:

`(x^2-x-1)(x^2-3x-1)=0`

We have already figured out when the left factor equals zero, so we now look at the right. We can solve the quadratic using the quadratic formula to get:
`x=(3+-sqrt13)/2`