https://in.mathworks.com/matlabcentral/fileexchange/54253-drawing-shape-function-of-quadratic-element-1d-or-beam-element

Certainly! Here's an example code for solving a quadratic equation using MATLAB:

```matlab

% Quadratic Equation Solver

% Input the coefficients of the quadratic equation

a = input('Enter the coefficient of x^2 (a): ');

b = input('Enter the coefficient of x (b): ');

c = input('Enter the constant term (c): ');

% Calculate discriminant

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

% Check the nature of roots based on the discriminant

if D > 0

disp('Roots are real and distinct.');

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

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

fprintf('Root 1 (x1): %.4f\n', x1);

fprintf('Root 2 (x2): %.4f\n', x2);

elseif D == 0

disp('Roots are real and equal.');

x = -b/(2*a);

fprintf('Root (x): %.4f\n', x);

else

disp('Roots are complex.');

realPart = -b/(2*a);

imagPart = sqrt(-D)/(2*a);

fprintf('Root 1: %.4f + %.4fi\n', realPart, imagPart);

fprintf('Root 2: %.4f - %.4fi\n', realPart, imagPart);

end

```

To use this code, simply copy and paste it into a MATLAB script file (e.g., `quadratic_solver.m`) and run the script. The program will prompt you to enter the coefficients of the quadratic equation (a, b, and c), and it will calculate and display the roots based on the nature of the roots (real and distinct, real and equal, or complex).

Please note that this code assumes the quadratic equation is in the form of `ax^2 + bx + c = 0`, and it uses the quadratic formula to compute the roots.

Hope it helps:credit AI