1. Divide the integer part of the square root into a section with two digits from the unit to the left, separated by apostrophes, and divided into several sections, indicating how many digits the square root is;
2. According to the number in the first paragraph on the left, find the number with the highest square root;
3. Subtract the square of the highest digit from the number in the first paragraph, and write the number in the second paragraph to the right of their difference to form the first remainder;
4. Multiply the obtained highest digit by 2 to try to divide the first remainder, and the obtained largest integer is used as the trial quotient;
5. Multiply the quotient by twice the highest digit of the quotient, and then multiply the quotient. If the product is less than or equal to the remainder, the quotient is the second digit of the square root; If the product is greater than the remainder, reduce the trial quotient and try again.
Note: If a positive number has a square root, then there must be two, and it is reciprocal. Obviously, if we know one of these two square roots, we can get the other square root in time according to the concept of reciprocal.
In a real number system, negative numbers cannot be squared. Only in a complex system can negative numbers be squared. The square root of a negative number is the pure imaginary number of a pair of yokes.
For example, the square root of-1 is I, and the square root of -9 is 3i, where I is an imaginary unit.
How to open extended data
Let A = X^3, and x is called the issuer. The issuer has a standard formula:
For example, A=5 is to find.
5 is between the third power of 1; 2 to the third power; Between (cubic of 1 = 1, cubic of 2 =8)
The initial value X0 can be 1. 1. 1.2, 1.3, 1.4, 1.5, 1.6,/kloc-0. For example, we take X0 = 1.9 according to the formula:
Step 1: x1=1.9+(5/1.92; - 1.9) 1/3= 1.7。
That is, 5/1.9×1.9 =1.385041.3850416-1.9 =- That is, take a 2-digit value, namely 1.7.
Step 2: x2 =1.7+(5/1.72; - 1.7) 1/3= 1.7 1。
That is, 5/1.7×1.7 =1.73010, 1.73- 1.7=0.03, 0.03×/kloc-. Take 3 digits, one digit more than the previous one.
Step 3: x3 =1.71+(5/1.712; - 1.7 1) 1/3= 1.709.
Step 4: x4 =1.709+(5/1.709 2; - 1.709) 1/3= 1.7099
This method can be adjusted automatically. The values in the first and third steps are too large, but the output value will be reduced automatically after calculation. Step two, step four, enter the value.
If it is too small, the output value will increase automatically. That is, 5 =1.7099 3;
Of course, the initial value X0 can also be 1. 1, 1.2, 1.3. . . Any one of 1.8 and 1.9 is x1=1.7 >; . Of course, we'd better use the intermediate value as the initial value in actual operation, that is, 1.5. 1.5+(5/ 1.5 & amp； sup2- 1.5) 1/3= 1.7。
Baidu Encyclopedia-Square Root Operation