What are the roots of #8x^3-7x^2-61x+6=0# ?

1 Answer
Sep 25, 2016

Answer:

The roots are:

#x_n = 1/24(7+2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3))#

for #n = 0, 1, 2#

Explanation:

#8x^3-7x^2-61x+6 = 0#

#color(white)()#
Discriminant

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=8#, #b=-7#, #c=-61# and #d=6#, so we find:

#Delta = 182329+7263392+8232-62208+368928 = 7760673#

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

#color(white)()#
Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=1728f(x)=13824x^3-12096x^2-105408x+10368#

#=(24x-7)^3-4539(24x-7)-21062#

#=t^3-4539t-21062#

where #t=(24x-7)#

#color(white)()#
Trigonometric substitution

Look for a substitution of the form #t = k cos theta# where #k# is chosen to cause the resulting equation to contain:

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

Let #t = k cos theta#

#0 = t^3-4539t-21062#

#color(white)(0) = k^3 cos^3 theta - 4539 k cos theta - 21062#

#color(white)(0) = (k^3/4) 4cos^3 theta - (1513 k) 3cos theta - 21062#

So we want:

#k^3/4 = 1513k#

Hence:

#k^2 = 4*1513#

So:

#k = +-2sqrt(1513)#

Let #k = 2sqrt(1513)#

Then #1513k = 3026sqrt(1513)#

Then:

#0 = (k^3/4) 4cos^3 theta - (1513 k) 3cos theta - 21062#

#color(white)(0) = 3026sqrt(1513)(4cos^3 theta - 3cos theta) - 21062#

#color(white)(0) = 3026sqrt(1513)(cos 3 theta) - 21062#

Hence:

#cos 3 theta = 21062/(3026 sqrt(1513)) = (10531 sqrt(1513))/2289169#

So:

#3 theta = +-cos^(-1)((10531 sqrt(1513))/2289169) + 2npi#

#theta = +- 1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3#

Hence:

#t_n = 2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3)#

with distinct values for #n = 0, 1, 2#

Then #x = 1/24(t + 7)#

So we have roots:

#x_n = 1/24(7+2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3))#

for #n = 0, 1, 2#