# Taylor expanding to leading order

I've had a lot of trouble finding a reduced form of the solutions here to the leading order:

$$\omega_{1,2}=-\frac{1}{2}(1+k+\epsilon) \pm \frac 12 \sqrt{(1+k+\epsilon)^2-4k\epsilon}$$

The textbook I'm using has the following assumptions: $k$ is of the order of 1, and $\epsilon$ is much smaller than 1. If we Taylor expand the above expression in powers of $\epsilon$ using the rule that $\sqrt{1+x}\approx 1+x/2$ we obtain the following eigenvalues:

$$\omega_1=-\frac {k}{k+1} \epsilon$$ $$\omega_2=-(k+1)$$ to leading order in $\epsilon$.

For the life of me I haven't been able to arrive at these solutions. What terms can I ignore in my expansions of the squared parenthesis in the root and the following Taylor expansion? Thank you.

$$\sqrt{(1+k+\epsilon)^2-4k\epsilon} = \sqrt{(1+k)^2 + 2(1+k)\epsilon + \epsilon^2 -4k\epsilon} = \sqrt{(1+k)^2 + 2(1-k)\epsilon + \epsilon^2} = (1+k)\sqrt{1 + \frac{2(1-k)\epsilon + \epsilon^2}{(1+k)^2}}$$
$$(1+k)(1 + \frac{(1-k)\epsilon}{(1+k)^2}) = (1 + k) + \frac{1-k}{1+k}\epsilon$$