What is the solution for a cubic equation? Thanks.....

1 Answer
Mar 20, 2016

Answer:

See explanation for a demonstration of Cardano's method...

Explanation:

There is a general formula and there are several methods.

My personal favourite method, which works well when a cubic equation has one Real root and two non-Real roots is Cardano's method.

Given a cubic equation:

#ax^3+bx^2+cx+d = 0#

If the coefficients are integers, then it may be useful to first multiply through by #3^3 a^2# to reduce the number of fractions we need to deal with.

#27a^3x^3 + 27a^2bx^2 + 27a^2cx+27a^2d=0#

#(3ax)^3+3b(3ax)^2+9ac(3ax)+27a^2d = 0#

Substitute #t = 3ax+b# to get:

#t^3+(9ac-3b^2)t+(27a^2d+2b^3-9abc) = 0#

Let #p = 9ac-3b^2# and #q = 27a^2d+2b^3-9abc#, so we have:

#t^3+pt+q = 0#

Next substitute #t = u + v# to get:

#u^3+v^3+(3uv+p)(u+v)+q = 0#

Then add the constraint: #v = -p/(3u)# to make #(3uv+p) = 0#

#u^3-p^3/(27u^3)+q = 0#

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

#27(u^3)^2+27q(u^3)-p^3 = 0#

Use the quadratic formula to find:

#u^3 = (-27q+-sqrt(729q^2+108p^3))/54#

#=(-27q+-3sqrt(81q^2+12p^3))/54#

The derivation was symmetric in #u# and #v#, so one of these roots can be used for #u^3# and the other for #v^3#.

If #81q^2+12p^3 >= 0# then the square root is Real and the Real root of the cubic #t^3+pt+q = 0# is:

#t_1 = root(3)((-27q+3sqrt(81q^2+12p^3))/54)+root(3)((-27q-3sqrt(81q^2+12p^3))/54)#

In this case, the Complex roots are given by:

#t_2 = omega root(3)((-27q+3sqrt(81q^2+12p^3))/54)+omega^2 root(3)((-27q-3sqrt(81q^2+12p^3))/54)#

#t_3 = omega^2 root(3)((-27q+3sqrt(81q^2+12p^3))/54)+omega root(3)((-27q-3sqrt(81q^2+12p^3))/54)#

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

If on the other hand #81q^2+12p^3 < 0# then the square roots are pure imaginary, the cube roots are of Complex numbers and all three roots are Real.

In this case to match the appropriate cube root #v# to #u#, we have to use the relation #v = -p/(3u)# to find:

#t_(k+1) = omega^k root(3)((-27q+3sqrt(81q^2+12p^3))/54) - p/(3 omega^k root(3)((-27q+3sqrt(81q^2+12p^3))/54))#

#k = 0, 1, 2#

From these vales of #t# we can find #x# by #x = (t-b)/(3a)#