One-dimensional quadratic equation and one-dimensional quadratic equation in C Language

One-dimensional quadratic equation and one-dimensional quadratic equation in C Language

When solving a quadratic equation using C language, we first agree that this equation is in the form of AX2 + BX + C = 0. Next we need to consider the parameter problem of this quadratic equation. We can discuss in three cases. When A = 0, B = 0, such A parameter cannot constitute an appropriate equation. When A = 0, B = 0, this equation is A one-dimensional equation. When A is less than 0, A quadratic equation is formed. The logic in these three cases can be easily implemented using the if statement. Next, we need to select the parameter type. Here we choose the float type, because the data stored in the computer is binary, there will be some errors when storing floating point numbers, we need to define an error precision, solve the problem within the error range. Finally, in the third case, We nested the if statement to solve the problem of different solutions when △is greater than or equal to 0 in a one-dimensional quadratic equation. The Code is as follows:

// Author: fl // date: 17.9.17 // function: solving the quadratic equation # define _ CRT_SECURE_NO_WARNINGS 1 # include
 
  
# Include
  
   
# Define disc 0.00000001 // error precision int main () {printf ("Please input the parameter a B C in the form of AX ^ 2 + BX + C = 0; \ n "); printf (" \ n "); float a = 0.0, B = 0.0, c = 0.0, d = 0.0, x1 = 0.0, x2 = 0.0; scanf ("% f", & a, & B, & c); printf ("\ n "); d = B * B-4 * a * c; // used to determine the number of real root numbers x1 = (-B) + sqrt (d)/(2 * ); // used to solve the real number root x2 = (-B)-sqrt (d)/(2 * a); if (a>-disc) & (a <disc) & (B>-disc) & (B <disc) // if the value of a and B is zero, cannot form an appropriate function {pri Ntf ("Enter reasonable parameters! \ N "); printf (" \ n ");} else if (a>-disc) & (a <disc )) & (B <-disc) | (B> disc) // if a is 0 and B is not 0, the parameters a, B, c Constitutes a one-dimensional equation {printf ("X = % f \ n", (-c)/B); printf ("\ n ");} else {if (d> disc) // d is greater than 0, and the equation has two different solutions {printf ("X1 = % f \ n", x1 ); printf ("X2 = % f \ n", x2); printf ("\ n");} else if (d>-disc) & (d <disc )) // d equals 0, and the equation has two identical solutions {printf ("X = X1 = X2 = % d \ n", x1); printf ("\ n ");} else // d is less than 0, and the equation has no solution {printf ("This equation has no real number root! \ N "); printf (" \ n ") ;}} return 0 ;}