It was the time in schools, where we were taught about finding square roots of numbers. For a long time, i found the technique very difficult, so always used to find some shortcut to get rid of the tedious method. Could not find anything useful, but while using my calculator I used to observe a pattern, something as follows :
SqRoot( x ) * SqRoot ( x+1 ) = Math.round ( (10 * x + 4 )/10 , 1 )
so like
SqRoot( 2 ) * SqRoot ( 3 ) = 2.4
SqRoot( 5 ) * SqRoot ( 6 ) = 5.4
SqRoot( 23 ) * SqRoot ( 24 ) = 23.4
The only useful thing i found about the above method was to use it for finding sq root of a number quickly ( But with only 1 decimal precision ) , when some number with easy square root is available nearby ( for example 4,16, 100 )
SqRoot(99) * SqRoot(100) = 99.4
So SqRoot ( 99) = 9.94
I am not sure, if this method is already explained somewhere on wiki.. or something similar is already available, or like might be just a part of some bigger equation. But i would do like to know and curious, that why something like this "pattern" occurs. And can it be extended more, to make square root methedology even simpler ?
Vishwas
This doesn't explain why the "pattern" occurs; it simply verifies that the pattern does occur. I hope it should shed some light on the matter anyway.
Let $\text{Math.truncate}(x, n)$ denote the truncation of $x$ to $n$ decimal digits. Then, the equation you propose (as I've understood it) is $$ \text{Math.truncate}(\sqrt{x} \cdot \sqrt{x + 1}, 1) = x + 0.4 $$
Assume that $x$ is a positive integer. Then, an equivalent way of writing the equation is $$ x + 0.4 \leq \sqrt{x} \cdot \sqrt{x + 1} < x + 0.5 $$
Because all numbers involved are positive, we can simplify this by squaring throughout the inequality: $$ (x + 0.4)^2 \leq x \cdot (x + 1) < (x + 0.5)^2 $$
This is of course equivalent to $$ x^2 + 0.8x + 0.16 \leq x^2 + x < x^2 + x + 0.25 $$
The inequality to the left ($x^2 + 0.8x + 0.16 \leq x^2 + x$) holds for positive integers $x$. Just subtract $x^2 + 0.8x$ from both sides and conclude that $0.16 \leq 0.2x$.
The inequality to the right ($x^2 + x < x^2 + x + 0.25$) obviously holds.
And so we have shown that $\text{Math.truncate}(\sqrt{x} \cdot \sqrt{x + 1}, 1) = x + 0.4$ is true for all positive integers $x$.