I want to show that taking the square root of a real number is backwards stable in the sense that for any $x \in \mathbb{R}$ there exists a $\tilde{x} \in \mathbb{R}$ s.t.
$$\frac{\vert x - \tilde{x} \vert}{\vert x \vert} = O(\epsilon_{mach}), ~~~~~~~~~~~~~ \hat{\sqrt{x}} = \sqrt{x}$$
where $\hat{\sqrt{ }}$ denotoes the algorithm to compute the square root.
Now there exists a $\epsilon > 0$ s.t. $\vert \epsilon \vert < \epsilon_{mach}$ where $\epsilon_{mach}$ is the machine epsilon s.t.
$$\hat{\sqrt{x}} = \sqrt{x}(1+\epsilon) = \sqrt{x(1+\epsilon)^2} = \sqrt{x(1+2 \epsilon + \epsilon^2)}$$
Now define $\delta := 2 \epsilon + \epsilon^2$, which then gives $\sqrt{x(1 + \delta)}$ and define $\tilde{x} := x(1 + \delta)$.
Now, my notes claim that
$$\vert \delta \vert \leq \vert \epsilon \vert^2 + 2 \vert \epsilon \vert \leq \vert \epsilon_{mach} \vert^2 + 2 \vert \epsilon_{mach} \vert = O(\epsilon_{mach})$$
I do not understand why the last equality holds. Why does $\vert \epsilon_{mach} \vert^2 + 2 \vert \epsilon_{mach} \vert = O(\epsilon_{mach})$?