How do you find all the zeros of #f(x)= 2x^3 - 6x^2 + 7x +9#?
1 Answer
Use Cardano's method to find Real zero:
#x_1 = 1/6(6+root(3)(648+6sqrt(11670))+root(3)(648-6sqrt(11670)))#
and related Complex zeros.
Explanation:
#f(x) = 2x^3-6x^2+7x+9#
Descriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 1764-2744+7776-8748-13608 = -15560#
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=4f(x)=8x^3-24x^2+28x+36#
#=(2x-2)^3+2(2x-2)+48#
#=t^3+2t+48#
where
Cardano's method
We want to solve:
#t^3+2t+48=0#
Let
Then:
#u^3+v^3+(3uv+2)(u+v)+48=0#
Add the constraint
#u^3-8/(27u^3)+48=0#
Multiply through by
#27(u^3)^2+1296(u^3)-8=0#
Use the quadratic formula to find:
#u^3=(-1296+-sqrt((1296)^2-4(27)(-8)))/(2*27)#
#=(1296+-sqrt(1679616+864))/54#
#=(1296+-sqrt(1680480))/54#
#=(1296+-12sqrt(11670))/54#
#=(648+-6(11670))/27#
Since this is Real and the derivation is symmetric in
#t_1=1/3(root(3)(648+6sqrt(11670))+root(3)(648-6sqrt(11670)))#
and related Complex roots:
#t_2=1/3(omega root(3)(648+6sqrt(11670))+omega^2 root(3)(648-6sqrt(11670)))#
#t_3=1/3(omega^2 root(3)(648+6sqrt(11670))+omega root(3)(648-6sqrt(11670)))#
where
Now
#x_1 = 1/6(6+root(3)(648+6sqrt(11670))+root(3)(648-6sqrt(11670)))#
#x_2 = 1/6(6+omega root(3)(648+6sqrt(11670))+omega^2 root(3)(648-6sqrt(11670)))#
#x_3 = 1/6(6+omega^2 root(3)(648+6sqrt(11670))+omega root(3)(648-6sqrt(11670)))#