Question

Please show me the MATLAB CODE for this problem Thank you

Write programs that determine solutions for polynomials (quadratics, et al)

1. a) Create a program that presumes the user-entered values (a,b,c) represent a valid quadratic equation and your initial program simply solves the equation and reports the real or imaginary roots.

2. b) Create a flowchart that describes all possible combinations of user entered data, e.g. perhaps a user enters (1, 0, -4) as (a, b, c) which is of course results in:

   22
   ax+bx+c=0 → x−4=0or=±x2

3. c) Create an efficient and user-friendly program that accepts any combination of constants (a,b,c) and returns the correct roots for the appropriate polynomial.

Answer 1

clear all ;
close all
clc

% the equation will be in the form
% a*x^2 + b*x + c =0
prompt = {'Enter a ','Enter b ','Enter c'};
dlg_title = 'Input';
num_lines = 1;
defaultans = {'1','6','4'};
answer = inputdlg(prompt,dlg_title,num_lines,defaultans);
a = str2double(answer{1});
b = str2double(answer{2});
c = str2double(answer{3});

D = b^2-4*a*c;
if D < 0
root1 = strcat(num2str(-b/2/a),'+',num2str(sqrt(abs(D))/2/a),'*I');
root2 = strcat(num2str(-b/2/a),'-',num2str(sqrt(abs(D))/2/a),'*I');
if b == 0
root1 = strcat(num2str(sqrt(abs(D))/2/a),'*I');
root2 = strcat('-',num2str(sqrt(abs(D))/2/a),'*I');
end
elseif D == 0
root1 = num2str(-b/2/a);
root2 = num2str(-b/2/a);
else
root1 = num2str(-b/2/a+sqrt(D)/2/a);
root2 = num2str(-b/2/a-sqrt(D)/2/a);
end

fprintf('%s = %s \n','root 1',root1)
fprintf('%s = %s \n','root 2',root2)

COMMENT DOWN FOR ANY QUERIES AND,

LEAVE A THUMBS UP IF THIS ANSWER HELPS YOU.