Solution of Non-linear equation by Bisection Method

Algorithm:

• Declare and initialize necessary variables like up_range, mid, low_range etc.
• Read the range( upper and lower) from user within which the root of the equation is to be calculated.
• If root lies within the range? if yes: go to step 4. if no: go to step 2
• Calculate the mid value of upper and lower range, mid = (upper+lower)/2
• Calculate the functional value at mid i.e. func(mid).
• If func(mid)*func(low_range) is less than zero, then replace upper range by mid else replace lower range by mid
• Display the no of iteration and root
• if func(mid) is very small? yes: go to step 9. No: go to step 4
• Display the value of most closest and accurate root.

Source Code:

#include
#include
//function that returns the functional value
float func(float x){
    return (pow(x,3)+5*pow(x,2)-7);
}
int main(){
    float up_range, low_range, mid;
    int i = 0; //no of iteration
    printf("Enter the range: ");
    scanf("%f%f",&up_range,&low_range);
    while(func(up_range)*func(low_range) > 0){ //repeatadly read until the range has root
        printf("\nThis range doesnot contains any root");
        printf("\nEnter again the range: ");
        scanf("%f%f",&up_range,&low_range);
    }
    do{
        mid = (up_range + low_range) / 2;
        if(func(low_range) * func(mid) < 0){ //if signs of mid and low_range is
            up_range = mid;                  //different, replace up_range by mid
        }else{ //else raplace, low_range by mid
            low_range = mid;
        }
        i++;
        printf("\nAt iteration: %d, root = %f",i,mid);
    }while(fabs(func(mid))> 0.0001);
    printf("\nThe root of the equation is %f", mid);
    return 0;
}