Recall that quadratic equations are of the form

Typically the solution is two real roots x1 and x2. Example: given 1, 3, and 2 for a, b, and c, solving for x yields -1 and -2.

Write a function SolveQuadratic that takes three coefficients a, b, and c, and solves for x. However, as you probably remember, in some cases there is only one solution for x, and in other cases there are no solutions for x. And when the discriminant is negative, solving for x yields complex roots --- which MATLAB supports. [ You can test this in MATLAB, Octave, or using the script window here in zybooks --- compute the square root of -1 and see what you get. Type this command: sqrt(-1) ]

In this exercise your function will have 3 output variables:

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

...

end

The output variables x1 and x2 denote the solution, while s is an information string about the solution. Here's the algorithm to follow for your function:

if a, b, and c are all 0

s = 'indeterminate';

elseif a and b are 0

s = 'none';

elseif a is 0

s = 'linear';

x1 = ...;

elseif the discriminant is 0

s = 'one real root';

x1 = ...;

elseif the discriminant is less than 0

s = 'complex roots';

x1 = ...; %% x1 = -b + ...

x2 = ...; %% x2 = -b - ...

else %% discriminant is positive:

s = 'real roots';

x1 = ...; %% x1 = -b + ...

x2 = ...; %% x2 = -b - ...

end

Recall that the discriminant is defined as b^2 - 4ac. To compute the square root of a value, use the MATLAB function sqrt. The syntax for if-then-else statements is accurately shown above, but you'll want to read about if statements, relational operators, and logical operators by skimming chapter 28 (CS 109) or chapter 11 (CS 110). You have unlimited submissions.

Because the function has 3 output variables, there's a special syntax necessary to call the function and capture all 3 results:

[x1, x2, s] = SolveQuadratic(1, 3, 2)