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

Jun 10, 2016

Use a numerical method to find approximate roots:

${x}_{1} \approx 12.9249$

${x}_{2} \approx 0.383152 + 0.64252 i$

${x}_{3} \approx 0.383152 - 0.64252 i$

${x}_{4} \approx - 0.691249$

#### Explanation:

$f \left(x\right) = {x}^{4} - 13 {x}^{3} + {x}^{2} - 5$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

$\pm 1$, $\pm 5$

None of these work, so $f \left(x\right)$ 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:

${p}_{0} = {\left(0.4 + 0.9 i\right)}^{0}$
${q}_{0} = {\left(0.4 + 0.9 i\right)}^{1}$
${r}_{0} = {\left(0.4 + 0.9 i\right)}^{2}$
${s}_{0} = {\left(0.4 + 0.9 i\right)}^{3}$

Then iterate using the formulas:

${p}_{i + 1} = {p}_{i} - f \frac{{p}_{i}}{\left({p}_{i} - {q}_{i}\right) \left({p}_{i} - {r}_{i}\right) \left({p}_{i} - {s}_{i}\right)}$

${q}_{i + 1} = {q}_{i} - f \frac{{q}_{i}}{\left({q}_{i} - {p}_{i + 1}\right) \left({q}_{i} - {r}_{i}\right) \left({q}_{i} - {s}_{i}\right)}$

${r}_{i + 1} = {r}_{i} - f \frac{{r}_{i}}{\left({r}_{i} - {p}_{i + 1}\right) \left({r}_{i} - {q}_{i + 1}\right) \left({r}_{i} - {s}_{i}\right)}$

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:

${x}_{1} \approx 12.9249$

${x}_{2} \approx 0.383152 + 0.64252 i$

${x}_{3} \approx 0.383152 - 0.64252 i$

${x}_{4} \approx - 0.691249$

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