What are the roots of #8x^3-7x^2-61x+6=0# ?
1 Answer
The roots are:
#x_n = 1/24(7+2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3))#
for
Explanation:
#8x^3-7x^2-61x+6 = 0#
Discriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 182329+7263392+8232-62208+368928 = 7760673#
Since
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
Trigonometric substitution
Look for a substitution of the form
#4cos^3 theta - 3 cos theta = cos 3 theta#
Let
#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
Then
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
Then
So we have roots:
#x_n = 1/24(7+2sqrt(1513) cos(1/3 cos^(-1)((10531 sqrt(1513))/2289169) + (2npi)/3))#
for