How do you find all the real and complex roots of #x^4 – 13x^3 + x^2 – 5 = 0#?

1 Answer
Jun 10, 2016

Answer:

Use a numerical method to find approximate roots:

#x_1 ~~ 12.9249#

#x_2 ~~ 0.383152 + 0.64252i#

#x_3 ~~ 0.383152 - 0.64252i#

#x_4 ~~ -0.691249#

Explanation:

#f(x) = x^4-13x^3+x^2-5#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-5# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible zeros are:

#+-1#, #+-5#

None of these work, so #f(x)# has no rational zeros.

It is possible to solve algebraically, but the solution is horribly messy.

Instead, find approximations to the zeros using a numerical method known as the Durand-Kerner method.

Suppose the #4# zeros are #p, q, r, s#.

Choose initial approximations:

#p_0 = (0.4+0.9i)^0#
#q_0 = (0.4+0.9i)^1#
#r_0 = (0.4+0.9i)^2#
#s_0 = (0.4+0.9i)^3#

Then iterate using the formulas:

#p_(i+1) = p_i-f(p_i)/((p_i-q_i)(p_i-r_i)(p_i-s_i))#

#q_(i+1) = q_i-f(q_i)/((q_i-p_(i+1))(q_i-r_i)(q_i-s_i))#

#r_(i+1) = r_i-f(r_i)/((r_i-p_(i+1))(r_i-q_(i+1))(r_i-s_i))#

#s_(i+1) = s_i-f(s_i)/((s_i-p_(i+1))(s_i-q_(i+1))(s_i-r_(i+1))#

Using this method and iterating a few times, we find approximations:

#x_1 ~~ 12.9249#

#x_2 ~~ 0.383152 + 0.64252i#

#x_3 ~~ 0.383152 - 0.64252i#

#x_4 ~~ -0.691249#

Here's a C++ program that I used...

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