How to solve #(3w-65)/(w^2+65)=w-7#?

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1 Answer
George C.
Mar 23, 2018

Real solution:

#w = 1/3(7+root(3)(3655+3sqrt(1770042))+root(3)(3655-3sqrt(1770042)))#

and related complex solutions.

Explanation:

Given:

#(3w-65)/(w^2+65) = w-7#

Multiply both sides by #w^2+65# to get:

#3w-65 = (w-7)(w^2+65)#

#color(white)(3w-65) = w^3-7w^2+65w-455#

Subtract #3w-65# from both sides to get:

#0 = w^3-7w^2+62w-390#

We can solve this cubic using Cardano's method.

Given:

#f(w) = w^3-7w^2+62w-390#

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #aw^3+bw^2+cw+d# is given by the formula:

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

In our example, #a=1#, #b=-7#, #c=62# and #d=-390#, so we find:

#Delta = 188356-953312-535080-4106700+3046680 = -2360056#

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(w)=27w^3-189w^2+1674w-10530#

#=(3w-7)^3+411(3w-7)-7310#

#=t^3+411t-7310#

where #t=(3w-7)#

Cardano's method

We want to solve:

#t^3+411t-7310=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv+137)(u+v)-7310=0#

Add the constraint #v=-137/u# to eliminate the #(u+v)# term and get:

#u^3-2571353/u^3-7310=0#

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

#(u^3)^2-7310(u^3)-2571353=0#

Use the quadratic formula to find:

#u^3=(7310+-sqrt((-7310)^2-4(1)(-2571353)))/(2*1)#

#=(7310+-sqrt(53436100+10285412))/2#

#=(7310+-sqrt(63721512))/2#

#=3655+-3sqrt(1770042)#

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)(3655+3sqrt(1770042))+root(3)(3655-3sqrt(1770042))#

and related Complex roots:

#t_2=omega root(3)(3655+3sqrt(1770042))+omega^2 root(3)(3655-3sqrt(1770042))#

#t_3=omega^2 root(3)(3655+3sqrt(1770042))+omega root(3)(3655-3sqrt(1770042))#

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

Now #w=1/3(7+t)#. So the roots of our original cubic are:

#w_1 = 1/3(7+root(3)(3655+3sqrt(1770042))+root(3)(3655-3sqrt(1770042)))#

#w_2 = 1/3(7+omega root(3)(3655+3sqrt(1770042))+omega^2 root(3)(3655-3sqrt(1770042)))#

#w_3 = 1/3(7+omega^2 root(3)(3655+3sqrt(1770042))+omega root(3)(3655-3sqrt(1770042)))#

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