Approximating $\sqrt x$ using Heron's method vs. Taylor's method

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let's say I want to approximate $f(x)=\sqrt x$ using an "old" method (whatever that means): where:

$y^2\leq x$

or

$y \leq \sqrt x$

$\sqrt x \approx \frac y2 + \frac{x}{2y}$

With remainder R such that

$\frac{-(x-y^2)^2}{8y^3} \leq R \leq 0$

I'm supposed to explain this method, using Taylor Polynomials.

A Taylor polynomial of degree 2 of $\sqrt x$, say evaluated at 100, and x = 101 would be

$\sqrt 101 \approx P_2(101)=10 + \frac{1}{20}(101-100)-\frac{1}{2000}(101-100)^2$

$\sqrt 101 \approx P_2(101)=\frac{201}{20}-\frac{1}{2000}=\frac{20099}{2000}\approx \sqrt101$

I skipped the calculations as it belongs to the previous question. Now, what I don't see is how these two are related. I see that the error term "R" is in some interval..

Any hints?