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

1 Answer
Mar 10, 2016

Answer:

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

Explanation:

Let #f(x) = 4x^3+2x-1#

By the rational root theorem, any roots of #4x^3+2x-1 = 0# will be expressible in the form #p/q# for integers #p# and #q# where #p# is a divisor of the constant term #-1# and #q# a divisor of the coefficient #4# of the leading term.

So the only possible rational roots are:

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

We find #f(x) != 0# for all of these values, so #f(x) = 0# has no rational roots.

Use Cardano's method, substituting #x = u + v# to get:

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

Add the constraint #v = -1/(6u)# to eliminate the term in #(u+v)# to get:

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

Multiply through by #54u^3# to get:

#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 #u# and #v#, we can take one of these roots as #u^3# and the other as #v^3# to find the Real root:

#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 #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.