How do you find all the zeros of f(x)=x^4-2x-1f(x)=x4−2x−1?
1 Answer
Reduce to solving two quadratics...
Explanation:
f(x) = x^4-2x-1f(x)=x4−2x−1
Since
x^4-2x-1x4−2x−1
= (x^2-ax+b)(x^2+ax+c)=(x2−ax+b)(x2+ax+c)
= x^4+(b+c-a^2)x^2+(b-c)ax+bc=x4+(b+c−a2)x2+(b−c)ax+bc
Equating coefficients and rearranging a little we get:
{ (b+c = a^2), (b-c = -2/a), (bc=-1) :}
Then:
(a^2)^2 = (b+c)^2 = (b-c)^2+4bc = (-2/a)^2-4 = 4/((a^2))-4
Multiply both ends by
(a^2)^3 + 4(a^2) - 4 = 0
Use Cardano's method to solve this cubic in
Let
Add the constraint
u^3-64/(27u^3)-4 = 0
Multiply through by
27(u^3)^2-108(u^3)-64 = 0
Use the quadratic formula to find:
u^3 = (108+-sqrt((-108)^2-4(27)(-64)))/(2*27)
=(108+-sqrt(11664+6912))/54
=(108+-sqrt(18576))/54
=(108+-12sqrt(129))/54
=(54+-6sqrt(129))/27
Hence Real root:
a^2 = 1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))
We can use the positive root as the value of
a = sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))
Then:
b = 1/2(a^2-2/a)
= 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))-1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))
c = 1/2(a^2+2/a)
= 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))+1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))
This leaves us with two quadratics to solve:
x^2-(sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))))x+( 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))-1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))) = 0
x^2+(sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))))x+( 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))+1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))) = 0
We can find the roots of these quadratics using the quadratic formula, which are the zeros of the original quartic
