# Logistic Equation

## Homework Statement

I'm having some difficulty understanding the last step of the following solved problem from my textbook:

We will compute the equilibrium points of $kP \left( 1- \frac{P}{N} \right) - C$.

$kP \left( 1- \frac{P}{N} \right) - C =0$

$-kP^2+kNP - CN =0$

The quadratic equations has solutions

$P= \frac{N}{2} \pm \sqrt{\frac{N^2}{4}-\frac{CN}{k}}$

## The Attempt at a Solution

So if we start off by

$-kP^2+kNP - CN =0$

and using the quadratic equation:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

I get:

$P= \frac{-kN \pm \sqrt{(kN)^2-4kC}}{2k}$

$= \frac{N}{2} \pm \frac{\sqrt{k^2N^2-4kC}}{2k}$

So, why is my answer different from the one in the textbook? And how can I end up with the solution in the book (that I've posted above)?

## Answers and Replies

$P= \frac{-kN \pm \sqrt{(kN)^2-4kC}}{2k}$

$= \frac{N}{2} \pm \frac{\sqrt{k^2N^2-4kC}}{2k}$

$2k = \sqrt{\text{(what?)}}$

$2k = \sqrt{\text{(what?)}}$

Oh, I see. I never thought of it. Thank you very much.