Ignore the last two digits of the number called out by the spectator and choose the memorised square which is just lower (or equal) to the remaining number. The corresponding square root is the first digit. of your answer
Now consider the last digit of the number called out by the spectator. If this is 0 or 5, then you immediately know that the last digit of your answer is also 0 or 5.
In all other cases, the last digit of the number called out will indicate two possible values for the last digit of the square root. For example, if the last digit is 9, then the square root may end in either 3 or 7.
/*Java program to find out square root of a given number * without using any Built-In Functions */ public class SquareRootDemo2 { public static void main(String[] args) { //Number for which square root is to be found double number = -12; //This method finds out the square root findSquareRoot(number); } /*This method finds out the square root without using any built-in functions and displays it */ public static void findSquareRoot(double number) { boolean isPositiveNumber = true; double g1; //if the number given is a 0 if(number==0) { System.out.println("Square root of "+number+" = "+0); } //If the number given is a -ve number else if(number<0) { number=-number; isPositiveNumber = false; } //Proceeding to find out square root of the number double squareRoot = number/2; do { g1=squareRoot; squareRoot = (g1 + (number/g1))/2; } while((g1-squareRoot)!=0); //Displays square root in the case of a positive number if(isPositiveNumber) { System.out.println("Square roots of "+number+" are "); System.out.println("+"+squareRoot); System.out.println("-"+squareRoot); } //Displays square root in the case of a -ve number else { System.out.println("Square roots of -"+number+" are "); System.out.println("+"+squareRoot+" i"); System.out.println("-"+squareRoot+" i"); } } }
The answer is squareroot r2+h2 squareroot radiusxradius + heightxheight
six on the square root of three
19
8
/*Java program to find out square root of a given number * without using any Built-In Functions */ public class SquareRootDemo2 { public static void main(String[] args) { //Number for which square root is to be found double number = -12; //This method finds out the square root findSquareRoot(number); } /*This method finds out the square root without using any built-in functions and displays it */ public static void findSquareRoot(double number) { boolean isPositiveNumber = true; double g1; //if the number given is a 0 if(number==0) { System.out.println("Square root of "+number+" = "+0); } //If the number given is a -ve number else if(number<0) { number=-number; isPositiveNumber = false; } //Proceeding to find out square root of the number double squareRoot = number/2; do { g1=squareRoot; squareRoot = (g1 + (number/g1))/2; } while((g1-squareRoot)!=0); //Displays square root in the case of a positive number if(isPositiveNumber) { System.out.println("Square roots of "+number+" are "); System.out.println("+"+squareRoot); System.out.println("-"+squareRoot); } //Displays square root in the case of a -ve number else { System.out.println("Square roots of -"+number+" are "); System.out.println("+"+squareRoot+" i"); System.out.println("-"+squareRoot+" i"); } } }
The answer is squareroot r2+h2 squareroot radiusxradius + heightxheight
six on the square root of three
2i , where i is the so-called imaginary number: squareroot of -1
The square root of 9 is 3 which is a rational number
you have to square the outside number and then multiply it by the number inside the squareroot sign.
Use the Pythagoras formula. In this case, squareroot(3002 + 4002) = 500.Use the Pythagoras formula. In this case, squareroot(3002 + 4002) = 500.Use the Pythagoras formula. In this case, squareroot(3002 + 4002) = 500.Use the Pythagoras formula. In this case, squareroot(3002 + 4002) = 500.
1.7724538509......
1000
19
8
11.1803398