How do you solve #x^3+2x^2 = 2# ?

1 Answer
Feb 4, 2017

Answer:

Use Cardano's method to find real root:

#x_1 = 1/3(-2+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))#

and related Complex roots.

Explanation:

First subtract #2# from both sides to get this equation into standard form:

#x^3+2x^2-2 = 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=1#, #b=2#, #c=0# and #d=-2#, so we find:

#Delta = 0+0+64-108+0 = -44#

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

#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=27(x^3+2x^2-2)#

#color(white)(0)=27x^3+54x^2-54#

#color(white)(0)=(3x+2)^3-12(3x+2)-38#

#color(white)(0)=t^3-12t-38#

where #t=(3x+2)#

#color(white)()#
Cardano's method

We want to solve:

#t^3-12t-38=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-4)(u+v)-38=0#

Add the constraint #v=4/u# to eliminate the #(u+v)# term and get:

#u^3+64/u^3-38=0#

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

#(u^3)^2-38(u^3)+64=0#

Use the quadratic formula to find:

#u^3=(38+-sqrt((-38)^2-4(1)(64)))/(2*1)#

#=(38+-sqrt(1444-256))/2#

#=(38+-sqrt(1188))/2#

#=(38+-6sqrt(33))/2#

#=19+-3sqrt(33)#

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)(19+3sqrt(33))+root(3)(19-3sqrt(33))#

and related Complex roots:

#t_2=omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33))#

#t_3=omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33))#

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

Now #x=1/3(-2+t)#. So the roots of our original cubic are:

#x_1 = 1/3(-2+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))#

#x_2 = 1/3(-2+omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33)))#

#x_3 = 1/3(-2+omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33)))#