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Square Root
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Don't Know correct place to ask this but is there a formula for working out a square root of any number
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There is no formula. But there is a procedure which you can apply repeatedly (iteration) until you get the square root to the accuracy required. You need an initial guess at the answer, but it doesn't need to be very accurate. It's called the Newton-Raphson Method and if you know how to program you can easily write some code to do it. For some details of the method see http://en.wikipedia.o...uare_root_of_a_number
There are other methods, but the above is probably the quickest since the number of correct decimal places doubles with each iteration.
See also http://www.sosmath.co...diff/der07/der07.html
There are other methods, but the above is probably the quickest since the number of correct decimal places doubles with each iteration.
See also http://www.sosmath.co...diff/der07/der07.html
yes, the method descibed is qiite straightforward and allows you to do calculations in your head.
To get a good approximation of square root of N:
Make a reasonable guess.
Divide your N by the guess.
Add the result to your guess.
Divide by 2.
Call your answer Y. That's a good approximation.
If you want you can repeat the process using Y as your next guess.
Example: square root of 18.
Guess 4
18÷4 = 4.5
4+4.5= 8.5
8.5 ÷ 2 = 4.25
This is pretty close to the true answer of 4.243
To get a good approximation of square root of N:
Make a reasonable guess.
Divide your N by the guess.
Add the result to your guess.
Divide by 2.
Call your answer Y. That's a good approximation.
If you want you can repeat the process using Y as your next guess.
Example: square root of 18.
Guess 4
18÷4 = 4.5
4+4.5= 8.5
8.5 ÷ 2 = 4.25
This is pretty close to the true answer of 4.243
There is a pseudo long division process. Hopefully you can follow my explanation which I've merged with an example.
Separate the digits of the number into pairs working left and right of the decimal point. The lead ‘pair’ may only be a single digit. e.g 985.96 becomes 9 85. 96 Set out the display as for long division but as yet there is no divisor.
Find the square root equal to or less than the 1st pair. E.g. √9 = 3. Put 3 as the start of the quotient (above the 9).
Multiply 3 by 20 = 60 and use this as the divisor for the next stage. Bring down the next pair of digits (85). From now on when ‘dividing’ you also have to allow for the square of the new part quotient to be accommodated. So, 85 ‘divides’ by 60 ONE time but the remainder is not 25. We need to accommodate 1² = 1; the remainder is 25 – 1 = 24 which is carried over to the next stage. Put 1 as the next digit of the quotient (above 85). The quotient now reads 31.
Multiply 31 by 20 = 620 and use this as the divisor for the next stage. Bring down the next pair of digits (96) but as there is a carry over, you are now ‘dividing’ 2496 by 620. This divides FOUR times and the remainder of 16 accommodates 4² = 16. So there is no final remainder. No further pairs of digits to bring down and when 4 is put as the next digit of the quotient (above 96) then answer reads 31.4.
This is a very basic example and hopefully you can follow my explanation. It’s a great arithmetic challenge but becomes number crunching with a vengeance when the square root of numbers containing seven or more digits is required.
Separate the digits of the number into pairs working left and right of the decimal point. The lead ‘pair’ may only be a single digit. e.g 985.96 becomes 9 85. 96 Set out the display as for long division but as yet there is no divisor.
Find the square root equal to or less than the 1st pair. E.g. √9 = 3. Put 3 as the start of the quotient (above the 9).
Multiply 3 by 20 = 60 and use this as the divisor for the next stage. Bring down the next pair of digits (85). From now on when ‘dividing’ you also have to allow for the square of the new part quotient to be accommodated. So, 85 ‘divides’ by 60 ONE time but the remainder is not 25. We need to accommodate 1² = 1; the remainder is 25 – 1 = 24 which is carried over to the next stage. Put 1 as the next digit of the quotient (above 85). The quotient now reads 31.
Multiply 31 by 20 = 620 and use this as the divisor for the next stage. Bring down the next pair of digits (96) but as there is a carry over, you are now ‘dividing’ 2496 by 620. This divides FOUR times and the remainder of 16 accommodates 4² = 16. So there is no final remainder. No further pairs of digits to bring down and when 4 is put as the next digit of the quotient (above 96) then answer reads 31.4.
This is a very basic example and hopefully you can follow my explanation. It’s a great arithmetic challenge but becomes number crunching with a vengeance when the square root of numbers containing seven or more digits is required.
That seems impressive viginge. I still prefer mine though.
To use you example of 985.96 I would guess 31 but to make the divsion in my head easier I'll use 30 as my guess.
985.96 divided by 30 = 32.86
Adding my guess of 30 gives 62.86
Divide by 2 and the answer is 31.43
To one decimal place that's 31.4
Pretty much spot on, and the same as your answer!
To use you example of 985.96 I would guess 31 but to make the divsion in my head easier I'll use 30 as my guess.
985.96 divided by 30 = 32.86
Adding my guess of 30 gives 62.86
Divide by 2 and the answer is 31.43
To one decimal place that's 31.4
Pretty much spot on, and the same as your answer!
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