How do you find all the real and complex roots of #f(x)=x^3+x-3#?

1 Answer
May 29, 2016

Answer:

Use Cardano's method to find the Real zero:

#x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))#

and related Complex zeros.

Explanation:

The discriminant #Delta# of a general cubic in the form #ax^3+bx^2+cx+d# is given by the formula:

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

So in our example, with #a=1#, #b=0#, #c=1# and #d=-3#, we find:

#Delta = 0-4-0-243+0 = -247 < 0#

Since #Delta < 0# this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

It can be solved using Cardano's method.

Let #x = u+v#

Then:

#u^3+v^3+(3uv+1)(u+v)-3 = 0#

Add the constraint #v = -1/(3u)# to eliminate the term in #(u+v)# and get:

#u^3-1/(27u^3)-3 = 0#

Multiply through by #27u^3# and rearrange a little to get:

#27(u^3)^2-81(u^3)-1 = 0#

Then use the quadratic formula to find:

#u^3 = (81+-sqrt(81^2-(4*27*-1)))/(2*27)#

#=(81+-sqrt(6561+108))/54#

#=(81+-sqrt(6669))/54#

#=(81+-3sqrt(741))/54#

Since this is Real and the derivation was symmetric in #u# and #v#m we can use one of these roots for #u^3# and the other for #v^3# to find the Real zero:

#x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))#

and Complex zeros:

#x_2 = 1/3(omega root(3)((81+3sqrt(741))/2)+omega^2 root(3)((81-3sqrt(741))/2))#

#x_3 = 1/3(omega^2 root(3)((81+3sqrt(741))/2)+omega root(3)((81-3sqrt(741))/2))#

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