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

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1 Answer
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)))