Write a c program to find the roots of a quadratic equation using if else?

Answer 1

/*Hello!! I'm aditya From Bangalore I have the solution of this program*/

#include <stdio.h>

#include <math.h>

main()

{

float a,b,c,x1,x2,delta=0;

printf("enter the value of a,b,c\n");

scanf("f%f",&a,&b,&c);

delta=((b*b)-(4ac));

if(a=0)

{

printf("the variables cannot form a quadratic equation\n");

}

else

{

x1=(-b)+(sqrt(((b*b)-(4ac))/2a))

x2=(-b)-(sqrt(((b*b)-(4ac))/2a))

}

if(delta=0)

{

printf("the roots are real and equal");

}

if(delta>0)

{

printf("the roots are real and distinct");

}

if(delta<0)

{

printf("the roots are imaginary");

x1=(-b)+(sqrt(((b*b)-((4ac))/(float)(2a))

x2=(-b)-(sqrt(((b*b)-((4ac))/(float)(2a))

}

printf("the roots of the equation are %2.2f\n",x1,x2,delta);

}

/*the program is written by using simple-if and if-else constructs*/

Answer 2

That depends what language you use. But basically, you need to write commands to:

• Tell the user what it is all about
• Ask for the coefficients a, b, and c, for the equation ax2 + bx + c = 0
• Check whether the discriminant (the expression b2 - 4ac) is positive, negative, or zero
• Depending on the result, use the quadratic formula to show the two real solutions, the single real solution, or the two complex solutions. If you are only interested in real solutions, in the latter case state that there are no real solutions.

Answer 3

lol no betch