Solve by using not Make sure your final draft of your final

Make sure your final draft of your final is all in one document (flowcharts, code, screen shots of output, calculations, etc.) For the final project, you will create a root finder calculator polynomials of degree zero, one, two and three. where a, b, c, z Element R. real numbers You will ask the user for the degree of the polynomial and then the coefficients and the constant values of the polynomial. Then you will return the location of the real roots. You don\'t have to acknowledge complex roots. You should print out your answer in just the black DOS window, no need for a file output here.

Solution

#include <stdio.h>
#include <string.h>
float solve_for_y(float a, float b, float c)
{
float Y = Y = -(b + c) / a;
return Y;
}
main()
{
float a, b, c, Y,determinant, root1,root2, realPart, imaginaryPart,n;
  
printf(\"\ Enter the order of equation\");
Scanf(\"%d\",&n);
  
if(n=1)
{
  
printf(\"\ Enter a linear equation in one variable of the form aY + b + c = 0 \");
printf(\"\ Enter the value of a, b, c respectively: \");
scanf(\"%f%f%f\", &a, &b, &c);
Y = solve_for_y(a, b, c);
printf(\"\ Solution is Y = %f\", Y);

}

if(n=2)

{

printf(\"Enter coefficients a, b and c: \");
scanf(\"%lf %lf %lf\",&a, &b, &c);

determinant = b*b-4*a*c;

// condition for real and different roots
if (determinant > 0)
{
// sqrt() function returns square root
root1 = (-b+sqrt(determinant))/(2*a);
root2 = (-b-sqrt(determinant))/(2*a);

printf(\"root1 = %.2lf and root2 = %.2lf\",root1 , root2);
}

//condition for real and equal roots
else if (determinant == 0)
{
root1 = root2 = -b/(2*a);

printf(\"root1 = root2 = %.2lf;\", root1);
}

// if roots are not real
else
{
realPart = -b/(2*a);
imaginaryPart = sqrt(-determinant)/(2*a);
printf(\"root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi\", realPart, imaginaryPart, realPart, imaginaryPart);
}
}

return 0;
}