MATLAB编程实现方程根的判断 【例】

% Purpose:

% This program solves for the roots of a quadratic equation

% of the form a*x^2 + b*x + c = 0. It calculates the answers

% regardless of the type of roots that the equation possesses.

%

% Define variables:

% a --Coefficient of x^2 term of equation

% b --Coefficient of x term of equation

% c --Constant term of equation

% discriminant --Discriminant of the equation

% imag_part --Imag part of equation (for complex roots)

% real_part --Real part of equation (for complex roots)

% x1 --First solution of equation (for real roots)

% x2 --Second solution of equation (for real roots)

% Prompt the user for the coefficients of the equation

disp ('This program solves for the roots of a quadratic ');

disp ('equation of the form A*X^2 + B*X + C = 0.');

a = input('Enter the coefficient A: ');

b = input('Enter the coefficient B: ');

c = input('Enter the coefficient C: ');

% Calculate discriminant

discriminant = b^2 - 4 * a * c;

% Solve for the roots, depending on the vlaue of the discriminant.

if discriminant > 0 % there are two real roots, so ...

x1 = (-b + sqrt(discriminant)) / (2*a);

x2 = (-b - sqrt(discriminant)) / (2*a);

disp('This equation has two real roots:');

fprintf('x1 = %f\n', x1);

fprintf('x2 = %f\n', x2);

elseif discriminant == 0 % there is one repeated root, so ...

x1 = ( -b ) / (2*a);

disp('This equation has two identical real roots:');

fprintf('x1 = x2 = %f\n', x1);

else % there are complex roots, so ...

real_part = (-b) / (2*a);

imag_part = sqrt( abs(discriminant)) / (2*a);

disp('This equation has complex roots:');

fprintf('x1 = %f + i %f \n',real_part, imag_part);

fprintf('x1 + %f - i %f \n', real_part, imag_part);

end