# Thread: Imaginary roots factoring help

1. $\displaystyle r^{2}-6r+6=0$
This might result in an $\displaystyle i$ root. I am not sure just stuck.

This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like $\displaystyle 2i$

2. Originally Posted by diddledabble
$\displaystyle r^{2}-6r+6=0$

This might result in an $\displaystyle i$ root. I am not sure just stuck.
$\displaystyle r^{2}-6r+6=0$

I would complete the square here

$\displaystyle (r^{2}-6r+9)+6-9=0$

$\displaystyle (r^{2}-6r+9)-3=0$

$\displaystyle (r-3)^{2}-3=0$

$\displaystyle (r-3)^{2}=3$

Now solving for r will determine the nature of the roots.

$\displaystyle (r-3)^{2}=3$

$\displaystyle r-3=\sqrt{3}$

$\displaystyle r=3\pm\sqrt{3}$

These look real to me.

3. Originally Posted by diddledabble
$\displaystyle r^{2}-6r+6=0$
This might result in an $\displaystyle i$ root. I am not sure just stuck.
[I take this to be the characteristic equation from a DE]

To find r, use the quadratic formula. You should get $\displaystyle r=3\pm\sqrt{3}$.

4. Originally Posted by diddledabble
This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like $\displaystyle 2i$
What is the matrix you're working with?

5. Originally Posted by diddledabble
$\displaystyle \mathbf{x}^{\prime}(t)=\begin{bmatrix}6&-3\\2&1\end{bmatrix}{\color{red}\mathbf{x}(t)}$ with initial value of $\displaystyle \mathbf{x}(0)=\begin{bmatrix}-10\\-6\end{bmatrix}$
I think you left out an important part (in red)

Let $\displaystyle \mathbf{A}=\begin{bmatrix}6&-3\\2&1\end{bmatrix}$.

Then to find eigenvalues, we set $\displaystyle \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0$.

In this case, it turns out that $\displaystyle \mathbf{A}-\lambda \mathbf{I}=\begin{bmatrix}6-\lambda & -3\\2&1-\lambda\end{bmatrix}$.

It now follows that $\displaystyle \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=\left(6-\lambda\right)\left(1-\lambda\right)-\left(2\right)\left(-3\right)=6-7\lambda+\lambda^2+6=\lambda^2-7\lambda+12$

Thus, $\displaystyle \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0\implies\lambda^2-7\lambda+12=0\implies\left(\lambda-4\right)\left(\lambda-3\right)=0\implies$ $\displaystyle \lambda=4$ or $\displaystyle \lambda=3$.