二次方程的求解（重写）---例题
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 dataCalculate the rootsWrite 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 cdiscriminant ← b^2 - 4 * a * cx1 ←( -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 equationdisp ('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 discriminantdiscriminant = b^2 - 4 * a * c;% Solve for the rootsx1 = ( -b + sqrt(discriminant) ) / ( 2 * a );x2 = ( -b - sqrt(discriminant) ) / ( 2 * a );% Display resultsdisp ('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 = -3x2 + 4x + 4 = 0 x= -2x2 + 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_root2This program solves for the roots of a quadratic equation of the form A*X^2 + B*X + C = 0. Enter the coefficient A: 1Enter the coefficient B: 5Enter the coefficient C: 6The roots of this equation are:x1 = (-2.000000) +i (0.000000)x2 = (-3.000000) +i (0.000000)>> calc_root2This program solves for the roots of a quadratic equation of the form A*X^2 + B*X + C = 0. Enter the coefficient A: 1Enter the coefficient B: 4Enter the coefficient C: 4The roots of this equation are:x1 = (-2.000000) +i (0.000000)x2 = (-2.000000) +i (0.000000)>> calc_root2This program solves for the roots of a quadratic equation of the form A*X^2 + B*X + C = 0. Enter the coefficient A: 1Enter the coefficient B: 2Enter the coefficient C: 5The 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。
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