二次方程的求解（重写）---例题

森之张卫东

二次方程的求解（重写）

复数的价值体现在它能使运算简化。

例如，我们在例3.2中已解决的二次方程的求解问题，但它根据判别式用到3个选项的选择结构，由于复数的出现，负数的平方根的处理将不困难。所以能够大大简化我们的计算。编写一个普通的程序，解一元二次方程的根，不管是什么类型的。用复变量，而不用选择结构。

1. 陈述问题

编写一个程序，解一元二次方程的根，不管有两个不同的实根，还是用两个相同的实根或两个不同复根。不需要检测判别式。

2. 定义输入输出

本程序所需要方程式

ax2 + bx + c = 0                   (3.1)

的三个系数a，b，c。输出是这个方程式的所有根。

3. 设计算法

这个程序从整体上可以分为三大步，即输入，计算，输出

Read the input data

Calculate the roots

Write out the roots

我们现在把每一步进行逐步细化。这时判别式的值对程序的执行过程不产生影响。伪代码如下：

Prompt the user for the coefficients a, b, and c.

Read a, b, and c

discriminant ← b^2 - 4 * a * c

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

x2 ←( -b -sqrt(discriminant) ) / (2*a)

Print 'x1 = ' , real(x1), ' + i ', imag(x1)

Print 'x2 = ' , real(x2), ' + i ', imag(x2)

4. 将算法转化为MATLAB语句

% Script file: calc_roots2.m

%

% 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.

%

% Record of revisions:

% Date Programmer Description of change

% ==== ========== =====================

% 12/06/98 S. J. Chapman Original code

%

% 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

% x1 -- First solution of equation

% x2 -- Second solution of equation

% 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

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

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

% Display results

disp ('The roots of this equation are:');

fprintf ('x1 = (%f) +i (%f)\n', real(x1), imag(x1));

fprintf ('x2 = (%f) +i (%f)\n', real(x2), imag(x2));

5.检测程序下一步，我们必须输入检测来检测程序。我们要有三组数据进行检测，其判别式分别大于0，等于0，小于0。根据方程式（3。1），用下面的方程式验证程序。

x2 + 5x + 6 = 0             x= -2, x = -3

x2 + 4x + 4 = 0             x= -2

x2 + 2x + 5 = 0             x= -1 ± 2i

我们把它们的系数分别输入程序，结果如下

>> calc_root2

This program solves for the roots of a quadratic

equation of the form A*X^2 + B*X + C = 0.

Enter the coefficient A: 1

Enter the coefficient B: 5

Enter the coefficient C: 6

The roots of this equation are:

x1 = (-2.000000) +i (0.000000)

x2 = (-3.000000) +i (0.000000)

>> calc_root2

This program solves for the roots of a quadratic

equation of the form A*X^2 + B*X + C = 0.

Enter the coefficient A: 1

Enter the coefficient B: 4

Enter the coefficient C: 4

The roots of this equation are:

x1 = (-2.000000) +i (0.000000)

x2 = (-2.000000) +i (0.000000)

>> calc_root2

This program solves for the roots of a quadratic

equation of the form A*X^2 + B*X + C = 0.

Enter the coefficient A: 1

Enter the coefficient B: 2

Enter the coefficient C: 5

The roots of this equation are:

x1 = (-1.000000) +i (2.000000)

x2 = (-1.000000) +i (-2.000000)

在三种不同的情况下，程序均给出了正确的结果。注意此程序与例3.2中的程序相比有多简单。复数数据的应用可大大简化我们的程序。