Блог

Когда я запускаю этот код, он извлекает странный символ, где должен быть ответ на математическую часть. Я попробовал значения a=1 b=4 c=40, и это не дает правильного ответа. Может кто-нибудь сказать мне, что я делаю не так? Я дважды проверил математику и почти уверен, что математическая часть в порядке.

 //Project by Abigail Alvarado
//Question 1

#include <stdio.h>
#include <math.h>

int main()
{
    //declaring variables

    float a, b, c; //These will be used in the quadratic equation that will give us lambda 1 and 2
    float d, root1, root2; //d is the quadratic equation determinant.
    float real, img;
    float y, theta;

    int t;
    t = 0;

    d = (b * b) - (4*a*c);

    /*D=d/dt or lambda; (y(0) and y'(0); ax^2   bx   c = 0 */


    //Asking for user input of coefficients
    printf("Enter the values for a, b, and c separated by a space: ");
    scanf("%f %f %f", amp;a, amp;b, amp;c);



    if(d < 0) //if d is less than 0; the roots are complex and different.

    {
        real = (-b/(2*a));
        img = sqrt(-d)/(2*a);
        printf("Roots are complex conjugates.n");

        printf("Roots of quadratic equation are: n Root1 = %.3f %.3fi n Root2 = %.3f-%.3fi", real, img, real, img);
    }

    else if(d==0) //if d==0; the roots are real and equal.
    {
        root1 = root2 = real;
        printf("Roots are repeating real numbers.n");
        printf("Roots of the quadratic equation are: n %.3f n %.3%f", root1, root2);
    }

    else if(d > 0) //if d is greater than 0; the roots are real and different.
    {
        real = (-b/(2*a));
        root1 = real   (sqrt(d)/(2*a));
        root2 = real - (sqrt(d)/(2*a));

        printf("Roots are non-repeating real numbers.n");

        printf("Roots of quadratic equation are: nRoot 1: %.3f nRoot 2: %.3f ", root1, root2);
    }



    //Asking user input of initial conditions; y(0)=? y'(0)=?

    float i1, i2; //initial conditions

    printf("nEnter two initial conditions separated by a space (y(0) and y'(0)): ");
    scanf("%f %f", amp;i1, amp;i2);

    //using initial conditions to find C and theta
    //Since t is equal to 0; e raised to (zero * any number) is equal to 1.

    printf("nUsing the first initial condition, C is equal to %.3f / (%.3f * theta). ", i1, (cos(img)));

    printf("nUsing the second initial condition, C is equal to %.3f / (%.3f * theta). ", i2, (cos(img)));



return (0);
}