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However did we manage before calculators? Let's think about that
for a moment.
Just taking square roots as an example, every time we use
Pythagoras to find the third side in a right-angled triangle we
need to perform a square root. If it's the square root of a number
like 4 that's easy : 2 . . plus or minus, if you like . . . but
what if we needed the square root of 40?
Obviously it isn't 20, but what is it?
Now we do have a technique, one that we use in school from time to
time, usually called 'trial and improvement', and it works like
this: get close to the right answer, and then try to get closer
still, being as systematic as possible, and continuing until we are
close enough for our purposes.
Well that's good, it's nice to have a solution method, but before
there were electronic calculators 'trial and improvement' could
involve quite a lot of work, so people were motivated to find
methods that used as little effort as possible.
Here's one of those methods, and your task is to decide how and why
SHIFT and click here to hear
or read the same explanation below the image.
Finding the square root of 40:
The answer is 6.324 . . . taking the method as far as three decimal
Notice the position of the decimal point and step away from there
in both directions, two digits at a time. Each of those pairs will
lead to one digit of the answer, and remember the zeros continue
Start on the left. The first pair is 40. Find the largest square
smaller than 40, that's 36. Subtract the 36 from the 40, which
leaves 4, and enter a 6 as the first digit of your answer.
Next take the 4 and 'bring down' the next pair of digits (00 as it
happens) to make 400. We are trying to decide the second digit of
our answer and we find it like this:
Use the 6 but double it (12) and make that ten times bigger (120).
Now find 'something'- a single digit, so that one hundred and
twenty 'something' times that same 'something' is as large as
possible but less than the 400.
123 times 3 makes 369, so three is the digit we want (124 times 4
would have been too big).
Subtract the 369 from the 400 (31) and 'bring down' the next pair
of digits (so we are now aiming for 3100).
The digits we have so far in the answer are 6 and 3. Double 63 and
then make it ten times bigger (1260). Use the same technique as
before: find one thousand two hundred and sixty 'something', times
'something', that gets as close as possible to 3100 without
Our third digit will be 2, 1262 times 2 is 2524, 1263 times 3 would
be too large.
Subtract to leave 576, bring down the next pair of digits, double
the digits you already have, that's 632, which doubles to make
1264, now look for twelve thousand, six hundred and forty
'something' times 'something' to come as close as possible to
57,600 without exceeding it. That 'something' is 4 (check it), and
so we continue until we have as many digits in our answer as we
think we need.
Your task is to find the next digit (the answer is on the Hint
page), check you can repeat the method for other square roots (use
your calculator to check your answers). Try to account for the
method and explain why it works?