The square root of a number x can be approximated by the repeated application of the following formula:

y(i) = (y(i-1)+ x/y(i-1)

where y(i) is the new estimate of the square root of x at iteration i, and y(i-1) is the estimate on the previous round.

Prompt the user to enter a positive number. Check whether the number is positive. If not, re-prompt the user to enter a positive number. Calculate the square root of the user's input using a C function that yields an accuracy of 5 decimal digits. That is, the new and the old estimate differ by less than 10^-5.

Use the following function prototype

 double root(double x);

The square root of a number x can be approximated by the repeated application of the following formula: y() = (y(i-1)+ x/y(i-1) where y(i) is the new estimate of the square root of x at iteration i, and y(i-1) is the estimate on the previous round. Prompt the user to enter a positive number. Check whether the number is positive. If not, re-prompt the user to enter a positive number. Calculate the square root of the user's input using a C function that yields an accuracy of 5 decimal digits. That is, the new and the old estimate differ by less than 10^-5. Use the following function prototype double root (double x); Sample Output Enter a positive number: -5 The number has to be positive. Enter a positive number: 7 The square root of 7 is 2.64575 301608.1549906 LAB ACTIVITY 16.25.1: Lab 4: P3(W): Square Root Approximation 0/4 main.c Load default template... 1 #include 2 #include 3 4 double root(double x); 5 6 int main() { /* Type your code here. */ 9 10 return 0; 11 } 12 13 double root(double x){ 14 15 // write your function here 16 17 }