**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:**

`/********************************************`

program: solution of non-linear equationbisection methodlanguage: CAuthor : Bibek SubediTribhuvan University, Nepal************************************************/#include<stdio.h>#include<math.h>//function that returns the functional valuefloat 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 iterationprintf("Enter the range: ");scanf("%f%f",&up_range,&low_range);while(func(up_range)*func(low_range) > 0){ //repeatadly read until the range has rootprintf("\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 isup_range = mid; //different, replace up_range by mid}else{ //else raplace, low_range by midlow_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;}

**Output: **

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