#include using namespace std;

double f(double x)
{
return (double) (x*x*x) + (x*x) - 1;
}

double g(double x) //equation of x
{
return (double) sqrt(1/(1+x));
}

int main()
{
double x0, a, b, xn=1;
int iter = 0;
x0 = 0;
a = f(x0);

// finding interval
while(1)
{
x0++;
b = f(x0);

if(a*b < 0)
{
a = x0 - 1;
b = x0;
break;
}

a = b;
}

cout<<"The interval I = ["< 0.0001)
{
x0 = xn;
xn = g(x0);
iter++;

cout<<" "< }

cout<<"\n\nThe root of the equation x^3 + x^2 - 1 = 0 is "<
return 0;
}

Convert this code to c language

imma come back to you?

answer; i do believe its (faster);

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