How to show this formula to get a square root of a number in "just few seconds" is true?

Solution 1:

To understand why this is a good approximation note that $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$.

So the formula is of the form $f(x) = f(y) + f'(y)(x-y)$. See Taylor's theorem for more details.

Solution 2:

For a purely algebraic and non-calculus derivation, use the fact that

$$x - y \; = \; (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}),$$

then replace $\;\sqrt{x} + \sqrt{y}\;$ with $\;\sqrt{y} + \sqrt{y} = 2\sqrt{y}\;$ (since we are assuming $\sqrt{x}$ is approximately equal to $\sqrt{y})$ to obtain the approximation

$$x - y \; \approx \; 2\sqrt{y} \left( \sqrt{x} - \sqrt{y} \right).$$

Now solve for $\sqrt{x}$ by dividing both sides by $2\sqrt{y}$ and adding $\sqrt{y}$ to both sides.