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How do you find the square root of 597? - Answers

The simplest method is to use the square root key on a calculator. But assuming you cannot do that: One alternative is the Newton-Raphson method. The process for using the method is described below, you can find out more about the rationale of the N-R method if you look for Newton's method on Wikipedia. Define f(x) = x2 - 597 Its derivative, f'(x) = 2x Make a guess at sqrt(597), say x0. Calculate xn+1 = xn + f(xn)/f'(xn) for n = 0, 1, 2 etc. Even with an outrageously high starting value, x0, of 30, the second iteration x2, has an error of around 2 in a hundred thousand! Another option is bracketing. Find two integers such that their squares bracket 597 242 = 576 < 597 < 625 = 252 so 24 < sqrt(597) < 25 Next find two numbers with 1 decimal place whose squares bracket 597 24.42 = 595.36 < 597 < 600.25 = 24.52 so 24.4 < sqrt(597) < 24.5 And so on, until you reach a satisfactory level of precision. Finally, there is a method that resembles long division but this browser is not suitable for describing that method.



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How do you find the square root of 597? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_find_the_square_root_of_597

The simplest method is to use the square root key on a calculator. But assuming you cannot do that: One alternative is the Newton-Raphson method. The process for using the method is described below, you can find out more about the rationale of the N-R method if you look for Newton's method on Wikipedia. Define f(x) = x2 - 597 Its derivative, f'(x) = 2x Make a guess at sqrt(597), say x0. Calculate xn+1 = xn + f(xn)/f'(xn) for n = 0, 1, 2 etc. Even with an outrageously high starting value, x0, of 30, the second iteration x2, has an error of around 2 in a hundred thousand! Another option is bracketing. Find two integers such that their squares bracket 597 242 = 576 < 597 < 625 = 252 so 24 < sqrt(597) < 25 Next find two numbers with 1 decimal place whose squares bracket 597 24.42 = 595.36 < 597 < 600.25 = 24.52 so 24.4 < sqrt(597) < 24.5 And so on, until you reach a satisfactory level of precision. Finally, there is a method that resembles long division but this browser is not suitable for describing that method.



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https://math.answers.com/math-and-arithmetic/How_do_you_find_the_square_root_of_597

How do you find the square root of 597? - Answers

The simplest method is to use the square root key on a calculator. But assuming you cannot do that: One alternative is the Newton-Raphson method. The process for using the method is described below, you can find out more about the rationale of the N-R method if you look for Newton's method on Wikipedia. Define f(x) = x2 - 597 Its derivative, f'(x) = 2x Make a guess at sqrt(597), say x0. Calculate xn+1 = xn + f(xn)/f'(xn) for n = 0, 1, 2 etc. Even with an outrageously high starting value, x0, of 30, the second iteration x2, has an error of around 2 in a hundred thousand! Another option is bracketing. Find two integers such that their squares bracket 597 242 = 576 < 597 < 625 = 252 so 24 < sqrt(597) < 25 Next find two numbers with 1 decimal place whose squares bracket 597 24.42 = 595.36 < 597 < 600.25 = 24.52 so 24.4 < sqrt(597) < 24.5 And so on, until you reach a satisfactory level of precision. Finally, there is a method that resembles long division but this browser is not suitable for describing that method.

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      The simplest method is to use the square root key on a calculator. But assuming you cannot do that: One alternative is the Newton-Raphson method. The process for using the method is described below, you can find out more about the rationale of the N-R method if you look for Newton's method on Wikipedia. Define f(x) = x2 - 597 Its derivative, f'(x) = 2x Make a guess at sqrt(597), say x0. Calculate xn+1 = xn + f(xn)/f'(xn) for n = 0, 1, 2 etc. Even with an outrageously high starting value, x0, of 30, the second iteration x2, has an error of around 2 in a hundred thousand! Another option is bracketing. Find two integers such that their squares bracket 597 242 = 576 < 597 < 625 = 252 so 24 < sqrt(597) < 25 Next find two numbers with 1 decimal place whose squares bracket 597 24.42 = 595.36 < 597 < 600.25 = 24.52 so 24.4 < sqrt(597) < 24.5 And so on, until you reach a satisfactory level of precision. Finally, there is a method that resembles long division but this browser is not suitable for describing that method.
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