1, integer square root step:
(1) The root number is separated by apostrophe every 2 digits from right to left; ?
(2) Find the first digit of the arithmetic square root in the first paragraph from the left; ?
(3) Subtract the square of the first digit from the first paragraph, and then write down the second paragraph of the root number as the first remainder; ?
(4) Multiply the first digit by 20, remove the first remainder, and take the integer part of the obtained quotient as the trial quotient (if the integer part is greater than or equal to 10, it is replaced by the left trial quotient 9, and if the first remainder is less than the product of the first digit multiplied by 20, the trial quotient 0 is obtained);
(5) Multiply the sum of the trial quotients by 20 times of the first digit. If the product obtained is greater than the remainder, you need to subtract the trial quotient from 1 and try again until the product is less than or equal to the remainder. This quotient is the second digit of the arithmetic square root; ?
(6) Continue to calculate other digits of the square root in the same way.
2, fractional part Kaiping method:
The square root of decimal can also be calculated by the general method of integer square root, but it is different to use apostrophe to segment. When segmenting, every two paragraphs are separated by apostrophes from the right of the decimal point.
If there is only one digit in the last paragraph after the decimal point, fill in a 0 to make up two digits, and then take the folk prescription of the integer as the step.
Second,
1. Manually calculate the square root (publisher) according to the sum of squares (sum of cubes) formula. In the past, the method of calculating the square root by hand, which was compulsory in junior high school textbooks, can be handled similarly by publishers.
2. Use dichotomy and inequality, such as finding the square root of 2.
1) 1^2<； 2 & lt2^2
2)( 1.4)^2<； 2 & lt( 1.5)^2
......
This method is very computationally intensive.
3. Differential approximation-because the error of this method is uncontrollable, the accuracy can be gradually improved by combining the former method, and the calculation amount is less than the former method.
4. The original Taylor expansion has a large amount of calculation and controllable error.
5. Deformed Taylor expansion, in terms of calculation method.
Reference link: mathematical resources