How to solve (3w-65)/(w^2+65)=w-7?
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1 Answer
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
3w-65 = (w-7)(w^2+65)
color(white)(3w-65) = w^3-7w^2+65w-455
Subtract
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 = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd
In our example,
Delta = 188356-953312-535080-4106700+3046680 = -2360056
Since
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
Cardano's method
We want to solve:
t^3+411t-7310=0
Let
Then:
u^3+v^3+3(uv+137)(u+v)-7310=0
Add the constraint
u^3-2571353/u^3-7310=0
Multiply through by
(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
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
Now
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)))