Why, using Newton's method for approximating roots, do distances have quadratic relationships?

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This MIT lecture defines $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}, \forall n\in\mathbb N$, where $x_1$ is a reasonable first guess for the root of the curve in the video. It then explains how $|x_a-x_{n+1}|^2\sim|x_a-x_n|$, where $x_a$ is the actual root. Could anyone explain why it is so?