What are the roots of #x^3+52x^2+1060x-4624 = 0#?

1 Answer
Jan 14, 2017

Answer:

The real root is:

#x = 2/3(-26+root(3)(21232+3sqrt(50275887))+root(3)(21232-3sqrt(50275887)))#

There are two related complex roots.

Explanation:

#f(x) = x^3+52x^2+1060x-4624#

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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=52#, #c=1060# and #d=-4624#, so we find:

#Delta = 3038214400-4764064000+2600685568-577297152-4587747840 = -4290209024#

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

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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(x)=27x^3+1404x^2+28620x-124848#

#=(3x+52)^3+1428(3x+52)-339712#

#=t^3+1428t-339712#

where #t=(3x+52)#

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Cardano's method

We want to solve:

#t^3+1428t-339712=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv+476)(u+v)-339712=0#

Add the constraint #v=-476/u# to eliminate the #(u+v)# term and get:

#u^3-107850176/u^3-339712=0#

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

#(u^3)^2-339712(u^3)-107850176=0#

Use the quadratic formula to find:

#u^3=(339712+-sqrt((-339712)^2-4(1)(-107850176)))/(2*1)#

#=(339712+-sqrt(115404242944+431400704))/2#

#=(339712+-sqrt(115835643648))/2#

#=(339712+-48sqrt(115835643648))/2#

#=169856+-24sqrt(115835643648)#

#=8(21232+-3sqrt(50275887))#

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=2root(3)(21232+3sqrt(50275887))+2root(3)(21232-3sqrt(50275887))#

and related Complex roots:

#t_2=2 omega root(3)(21232+3sqrt(50275887))+2 omega^2 root(3)(21232-3sqrt(50275887))#

#t_3=2 omega^2 root(3)(21232+3sqrt(50275887))+2 omega root(3)(21232-3sqrt(50275887))#

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

Now #x=1/3(-52+t)#. So the roots of our original cubic are:

#x_1 = 2/3(-26+root(3)(21232+3sqrt(50275887))+root(3)(21232-3sqrt(50275887)))#

#x_2 = 2/3(-26+omega root(3)(21232+3sqrt(50275887))+omega^2 root(3)(21232-3sqrt(50275887)))#

#x_3 = 2/3(-26+omega^2 root(3)(21232+3sqrt(50275887))+omega root(3)(21232-3sqrt(50275887)))#