What are the roots of #-9x^3+8x^2-2x+1 = 0# ?

1 Answer
George C.
Feb 2, 2018

This has roots:

#x_n = 1/27(8+omega^(n-1) root(3)((1915+135 sqrt(201))/2)+omega^(1-n) root(3)((1915-135 sqrt(201))/2))#

for #n = 1, 2, 3#

where #omega=-1/2+sqrt(3)/2i#

Explanation:

Given:

#f(x) = -9x^3+8x^2-2x+1#

Note that the pattern of signs of the coefficients of #f(x)# is #- + - +#. With #3# changes of signs, Descartes' Rule of Signs allows us to deduce that #f(x)# has #1# or #3# positive real zeros.

The pattern of signs of #f(-x)# is #+ + + +#. With no change of signs, we can deduce that #f(x)# has no negative real zeros.

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #1# and #q# a divisor of the coefficient #-9# of the leading term.

In combination with our previous observation that any real zeros are positive, we can deduce that the only possible rational zeros are:

#1/9, 1/3, 1#

None of these work, so #f(x)# has no rational zeros.

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=-9#, #b=8#, #c=-2# and #d=1#, so we find:

#Delta = 256-288-2048-2187+2592 = -1675#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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=-2187f(x)=19683x^3-17496x^2+4374x-2187#

#=(27x-8)^3-30(27x-8)-1915#

#=t^3-30t-1915#

where #t=(27x-8)#

Cardano's method

We want to solve:

#t^3-30t-1915=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-10)(u+v)-1915=0#

Add the constraint #v=10/u# to eliminate the #(u+v)# term and get:

#u^3+1000/u^3-1915=0#

Multiply through by #u^3# and rearrange slightly to get:

#(u^3)^2-1915(u^3)+1000=0#

Use the quadratic formula to find:

#u^3=(1915+-sqrt((-1915)^2-4(1)(1000)))/(2*1)#

#=(1915+-sqrt(3667225-4000))/2#

#=(1915+-sqrt(3663225))/2#

#=(1915+-135 sqrt(201))/2#

Since this is Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find Real root:

#t_1=root(3)((1915+135 sqrt(201))/2)+root(3)((1915-135 sqrt(201))/2)#

and related Complex roots:

#t_2=omega root(3)((1915+135 sqrt(201))/2)+omega^2 root(3)((1915-135 sqrt(201))/2)#

#t_3=omega^2 root(3)((1915+135 sqrt(201))/2)+omega root(3)((1915-135 sqrt(201))/2)#

where #omega=-1/2+135 sqrt(201)/2i# is the primitive Complex cube root of #1#.

Now #x=1/27(8+t)#. So the roots of our original cubic are:

#x_1 = 1/27(8+root(3)((1915+135 sqrt(201))/2)+root(3)((1915-135 sqrt(201))/2))#

#x_2 = 1/27(8+omega root(3)((1915+135 sqrt(201))/2)+omega^2 root(3)((1915-135 sqrt(201))/2))#

#x_3 = 1/27(8+omega^2 root(3)((1915+135 sqrt(201))/2)+omega root(3)((1915-135 sqrt(201))/2))#

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