# How do you solve #0 = 4x^3 + 2x-1#?

##### 1 Answer

#### Answer:

Check there are no rational roots, then use Cardano's method to find the Real and two Complex roots.

#### Explanation:

Let

By the rational root theorem, any roots of

So the only possible rational roots are:

#+-1/4# ,#+-1/2# ,#+-1#

We find

Use Cardano's method, substituting

#4u^3 + 4v^3 + 2(6uv+1)(u+v) - 1 = 0#

Add the constraint

#0 = 4u^3 - 4/(6u)^3 - 1 = 4u^3 - 1/(54u^3) - 1#

Multiply through by

#216(u^3)^2-54(u^3)-1 = 0#

Use the quadratic formula to find:

#u^3 = (54+-sqrt(54^2+4*216))/(2*216)#

#=(54+-sqrt(3780))/432#

#=(54+-6sqrt(105))/432#

#=(27+-3sqrt(105))/216#

Due to the symmetry of the derivation in

#x_1 = 1/6(root(3)(27+3sqrt(105))+root(3)(27-3sqrt(105)))#

and Complex roots:

#x_2 = 1/6(omega root(3)(27+3sqrt(105))+omega^2 root(3)(27-3sqrt(105)))#

#x_3 = 1/6(omega^2 root(3)(27+3sqrt(105))+omega root(3)(27-3sqrt(105)))#

where