How do you find all the zeros of # f(x) = -9x^3 + x^2 + 4x - 3#?
1 Answer
Use Cardano's method to find Real zero:
#x_1 = 1/27(1+root(3)((6235+27sqrt(46221))/2)+root(3)((6235-27sqrt(46221))/2))#
and related Complex zeros.
Explanation:
#f(x) = -9x^3+x^2+4x-3#
Descriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 16+2304+12-19683+1944 = -15407#
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=-2187f(x)=19683x^3-2187x^2-8748x+6561#
#=(27x-1)^3-327(27x-1)+6235#
#=t^3-327t+6235#
where
Cardano's method
We want to solve:
#t^3-327t+6235=0#
Let
Then:
#u^3+v^3+3(uv-109)(u+v)+6235=0#
Add the constraint
#u^3+1295029/u^3+6235=0#
Multiply through by
#(u^3)^2+6235(u^3)+1295029=0#
Use the quadratic formula to find:
#u^3=(-6235+-sqrt((6235)^2-4(1)(1295029)))/(2*1)#
#=(6235+-sqrt(38875225-5180116))/2#
#=(6235+-sqrt(33695109))/2#
#=(6235+-27sqrt(46221))/2#
Since this is Real and the derivation is symmetric in
#t_1=root(3)((6235+27sqrt(46221))/2)+root(3)((6235-27sqrt(46221))/2)#
and related Complex roots:
#t_2=omega root(3)((6235+27sqrt(46221))/2)+omega^2 root(3)((6235-27sqrt(46221))/2)#
#t_3=omega^2 root(3)((6235+27sqrt(46221))/2)+omega root(3)((6235-27sqrt(46221))/2)#
where
Now
#x_1 = 1/27(1+root(3)((6235+27sqrt(46221))/2)+root(3)((6235-27sqrt(46221))/2))#
#x_2 = 1/27(1+omega root(3)((6235+27sqrt(46221))/2)+omega^2 root(3)((6235-27sqrt(46221))/2))#
#x_3 = 1/27(1+omega^2 root(3)((6235+27sqrt(46221))/2)+omega root(3)((6235-27sqrt(46221))/2))#