The roots of a quadratic equation ax^2 + bx + c = 0 are given by the formula: x

Question

The roots of a quadratic equation ax^2 + bx + c = 0 are given by the formula: x = - b plusminus Squareroot b^2 - 4 ac/2a Write a function that takes as input the coefficients a, b, c and returns as output the roots x_1, x_2 of the quadratic equation. Let x_1 be the smaller of the two roots. If a is zero, print a warning message informing the user of the situation. Then, solve the linear equation for the single root x_1 = -c/b and set x_2 = NaN. If b^7 - 4 ac is negative, print a warning message informing the user of the situation Then, solve the equation anyways to find and return the complex roots If b^2 - 4ac is zero, set x_2 = x_1 = -b/2a If any of a. b, c are not scalar numbers, print an error message informing the user of the situation and set x_1 = x_2 = NaN Test your function extensively with an adversarial mindset Try to break it (make it crash or behave in undefined or erratic ways). Every time you break it, fix it. Your goal is to make the function able to handle any possible input, and to fail gracefully when the inputs are invalid Useful functions for this include isnumeric, isscalar, isreal, isinf, isnan.

Explanation / Answer

function [x1 x2] = rootsOfQuadratic(a, b, c)

if ~isscalar(a) || ~isscalar(b) || ~isscalar(c)

x1 = NaN;

x2 = NaN;

return;

end

if a == 0

fprintf('The equation is not quadratic. ');

x1 = -c / b;

x2 = NaN;

return;

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

if det < 0

fprintf('The roots are imaginary. ');

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

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

return;

end

if det == 0

x2 = -b / (2 * a);

x1 = x2;

return;

end

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

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

end

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