How do you calculate calculate square root?

There is a long and complicated way to do it, but there is another simpler way called successive aproximation. To find the square root (S) of X, the equation is S 2 X. Make a guess for S, call your guess G, and there must be a correction to G.

You have (G+c) 2 X. Multiply out the brackect, G 2 2Gc + c 2 X. Now I f your guess is reasonably good, you can ignore tiny c 2 so you get G 2 2Gc = X and you can sort this equation out to find c: c = (X-G 2 )/(2G).

Using this calculated correction c, find G+c which is now a better guess than before. So call this G+c you new value of G and go round again. Keep going round this loop until you are happy with the accuracy Another method (beloved by those brought up on mechanical calculators) relies on the fact that the sum of the first n odd numbers is n 2 (1+3+5+7=4 2 ).

If there is a decimal point, first multiply by 100 enough times to get rid of it. Thus for 1.69, multiply by 100 once to get 169. For some numbers you might want to extend the result to more places of decimals to the right of the decimal point, and in that case also multiply by 100 for each extra place of decimals you want to have.

Then subtract 1, 3, 5.... until there is nothing left, counting how many subtractions you make. So with 169, take away 1 (leaving 168), 3 (165), 5 (160), 7 (153), 9 (144), 11 (133), 13 (121), 15 (105), 17 (88), 19 (69), 21 (48), 23 (25), 25 (0). That's 13 subtractions Next divide by 10 the same number of times you multiplied by 100 in the first place.

That was once. So the answer is 1.3 . I repeat, if you want more decimal places of accuracy, just multiply by more 100s in the first place.

If the very last subtraction doesn't leave you with a zero, see if what is left and decide whether you want to add a small compensatory number for the last term or not, depending on the size of what's left. It will make only a tiny difference in most cases if you have allowed a sufficiently large number of places.

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