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

1 Answer
George C.
Aug 12, 2016

Use a trigonometric method to find roots:

#x_k = (4sqrt(3))/3 cos(1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3) " "# for #k = 0, 1, 2.#

Explanation:

#color(white)()#

#f(x) = x^3-4x-2#

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Descriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=0#, #c=-4# and #d=-2#, so we find:

#Delta = 0+256+0-108+0 = 148#

Since #Delta > 0# this cubic has #3# Real zeros.

Trigonometric solution

Use a substitution of the form #x = k cos theta#, choosing #k# so that the resulting expression contains #4cos^3 theta - 3 cos theta = cos 3 theta#.

Let #k = (4sqrt(3))/3#

Then:

#0 = x^3-4x-2#

#= k^3 cos^3 theta - 4k cos theta - 2#

#= (64 sqrt(3))/9 cos^3 theta - (16 sqrt(3))/3 cos theta - 2#

#= (16 sqrt(3))/9 (4 cos^3 theta - 3 cos theta) - 2#

#= (16 sqrt(3))/9 cos 3 theta - 2#

Hence:

#cos 3 theta = 2* 9/(16 sqrt(3)) = (3 sqrt(3))/8#

So:

#3 theta = +-cos^(-1)((3 sqrt(3))/8) + 2kpi#

#theta = +-1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3#

#cos theta = cos(1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3)#

Hence distinct roots:

#x_k = (4sqrt(3))/3 cos(1/3 cos^(-1)((3 sqrt(3))/8) + (2kpi)/3) " "# for #k = 0, 1, 2.#

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