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Explanation of irrational root theorem and imaginary root theorem

The Irrational Root Theorem

The Irrational Root Theorem says if $ a + \sqrt{b}$ is also a root of observed polynomial. In other words, irrational roots come in conjugate pairs.

Irrational Root Theorem

Example 1. Find the rational and irrational roots of the following polynomial equation.

$ x^3 + x^2 – 3x – 3 = 0$

If this equation has imaginary roots, by the Imaginary Root Theorem,  must divide 5.

$ a^2 + b^2 \in \{ 1, 5\}$

Now we have to think all the ways these numbers can be written as the sum of two squares of complex numbers.

First, for the number 1: $ 1 = 0 + 1 = 0 + (\pm 1)$ . This means that for this case $ a + bi \in \{\pm i \}$.

Second, for the number 5: $ 5 = 1 + 4$ and $ 5 = 4 + 1$. For this case $ a + bi \in \{ \pm1\pm2i, \pm2i \pm1\}$.

Now we have 9 possible solutions:

$ a + bi \in \{ \pm1\pm2i, \pm2i \pm1\}$

To eliminate few of these solution we’ll use the noted addition to the Imaginary Root Theorem.

We’ll use $ k = – 1$. $ (a – (- 1))^2 + b^2$ must divide $ f(- 1)$.

$ a + 1 + b^2$ must be divisible by 2. We’ll use this test for every possible solution. In the end we are left with:

$ a + bi \in \{ \pm i, – 2 \pm i \}$

$ x_1 = – 2 + i$, $ x_2 = – 2 – i$

After finding these two solutions we have a quadratic equation left to solve. This will lead us to the remaining two solutions to the given equation:

$ x_3 = \frac{1}{2} (- 1 + i\sqrt{3} )$, $ x_4 = \frac{1}{2} (- 1 – i\sqrt{3} )$

Imaginary root theorem

As an addition to this theorem, for every whole number $ k, (a – k)^2 + b^2$  divides $ f(k)$.

Example 2. Find all the roots of the following polynomial equation:

$ x^4 + 5x^3 + 10x^2 + 9x + 5 = 0$

If this equation has imaginary roots, by the Imaginary Root Theorem, $ a^2 + b^2$ must divide 5.

$ a^2 + b^2 \in \{ 1, 5 \}$

Now we have to think all the ways these numbers can be written as the sum of two squares of complex numbers.

First, for the number 1: $ 1 = 0 + 1 = 0 + (\pm 1)^2$ . This means that for this case $ a + bi \in \{ \pm 1 \}$.

Second, for the number 5: $ 5 = 1 + 4$ and $ 5 = 4 + 1$. For this case $ a + bi \in \{ \pm 1 \pm 2i, \pm 2i \pm 1 \}$

Now we have 9 possible solutions:

$ a + bi \in \{ \pm i, \pm 1 \pm 2i, \pm 2i \pm 1 \}$

To eliminate few of these solution we’ll use the noted addition to the Imaginary Root Theorem.

We’ll use $ k = – 1$. $ (a – (- 1))^2 + b^2$ must divide $ f(- 1)$.

$ a + 1 + b^2$ must be divisible by 2. We’ll use this test for every possible solution. In the end we are left with:

$ a + bi \in \{ \pm i, – 2 \pm i \}$

$ x_1 = -2 + i$, $ x_2 = – 2 – i$

After finding these two solutions we have a quadratic equation left to solve. This will lead us to the remaining two solutions to the given equation:

$ x_3 = \frac{1}{2}(- 1 + i\sqrt{3})$, $ x_4 = \frac{1}{2}(- 1 – i\sqrt{3})$