What are the roots of #t^3+2t^2+4t+6 = 0# ?

1 Answer
George C.
Jan 30, 2018

Use Cardano's method to find roots:

#t_n = 1/3(-2+omega^(n-1) root(3)(53+9sqrt(41))+omega^(1-n) root(3)(53-9sqrt(41)))#

for #n = 1, 2, 3#

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

Explanation:

Given:

#t^3+2t^2+4t+6 = 0#

By the rational roots theorem, any rational roots of this cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #6# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational roots are:

#+-1, +-2, +-3, +-6#

None of these work, so the cubic has no rational zeros.

So let's use a general solution method.

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #f(t) = at^3+bt^2+ct+d# is given by the formula:

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

In our example, #a=1#, #b=2#, #c=4# and #d=6#, so we find:

#Delta = 64-256-192-972+864 = -492#

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=27f(t)=27t^3+54t^2+108t+162#

#=(3t+2)^3+24(3t+2)+106#

#=x^3+24x+106#

where #x=(3t+2)#

Cardano's method

Let #x=u+v#

Then:

#u^3+v^3+3(uv+8)(u+v)+106=0#

Add the constraint #v=-8/u# to eliminate the #(u+v)# term and get:

#u^3-512/u^3+106=0#

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

#(u^3)^2+106(u^3)-512=0#

Use the quadratic formula to find:

#u^3=(-106+-sqrt((106)^2-4(1)(-512)))/(2*1)#

#=(-106+-sqrt(11236+2048))/2#

#=-53+-9sqrt(41)#

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:

#x_1 = root(3)(-53+9sqrt(41))+root(3)(-53-9sqrt(41))#

and related Complex roots:

#x_2 = omega root(3)(-53+9sqrt(41))+omega^2 root(3)(-53-9sqrt(41))#

#x_3 = omega^2 root(3)(-53+9sqrt(41))+omega root(3)(-53-9sqrt(41))#

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

Then using #t = 1/3(-2+x)# we find the roots of our original cubic:

#t_1 = 1/3(-2+root(3)(-53+9sqrt(41))+root(3)(-53-9sqrt(41)))#

#t_2 = 1/3(-2+omega root(3)(-53+9sqrt(41))+omega^2 root(3)(-53-9sqrt(41)))#

#t_3 = 1/3(-2+omega^2 root(3)(-53+9sqrt(41))+omega root(3)(-53-9sqrt(41)))#

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