Approximation of square roots Recently, I've seen a YouTube video where they approximate square roots real quick.
They use this approximation :
$$\sqrt{x} \approx \lfloor \sqrt{x} \rfloor+\dfrac{x-(\lfloor \sqrt{x} \rfloor)^2}{2\lfloor \sqrt{x} \rfloor}$$
I want to know the math behind this approximation. Can someone help me out?
 A: Source: How to find square roots
Imagine you have a square of area $x$. The first estimate of the side length is about $\lfloor\sqrt x\rfloor$, sounds reasonable?
But then that's an underestimation of that side length, because the estimated $\lfloor\sqrt x\rfloor\times\lfloor\sqrt x\rfloor$ square $\square$ is missing an L-shaped region $\blacksquare$, whose area is $\left(x-\lfloor\sqrt x\rfloor^2\right)$!
$$\left.\begin{array}{ccccccc}
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\square&\square&\square&\square&\square&\square&\blacksquare\\
\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare\\
\end{array}\quad\right\}\sqrt{x}=\lfloor\sqrt x\rfloor^2+\ldots?$$
Then the next estimate is to pretend that the L-shaped region is just two $\lfloor\sqrt x\rfloor\times\left( \frac{x-\lfloor\sqrt x\rfloor^2}{2\lfloor\sqrt x\rfloor}\right)$ rectangles. This give an over-estimated side length of the square
$$\begin{align*}
l &= \sqrt{x}\\
&\approx \lfloor\sqrt x\rfloor+\frac{x-\lfloor\sqrt x\rfloor^2}{2\lfloor\sqrt x\rfloor}
\end{align*}$$
A: Let, $n \in N $
Suppose, we want to evaluate $\sqrt{n^2+c}$
$$\implies \sqrt{n^2(1+\dfrac{c}{n^2})}$$
$$\implies n \sqrt{1+\dfrac{c}{n^2}}$$
Now, we shall use the binomial expansion for $\sqrt{1+\dfrac{c}{n^2}}$
We know, $\sqrt{1+\dfrac{c}{n^2}}=1+\dfrac{c}{2n^2}-\dfrac{c^2}{8n^4}+\cdots$
So, $n \sqrt{1+\dfrac{c}{n^2}}=n+\dfrac{c}{2n}-\dfrac{c^2}{8n^3}+ \cdots$
Therefore, we can say, $\sqrt{n^2+c}=n+\dfrac{c}{2n}+\cdots$
From there we get the approximation, $\sqrt{x} \approx \lfloor x \rfloor + \dfrac{x-(\lfloor \sqrt{x} \rfloor)^2}{2\lfloor \sqrt{x} \rfloor}$
