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bsd on 8 Nov 2011
Hai,
BSD

Walter Roberson on 8 Nov 2011
roots(). Or if you have the symbolic toolbox, solve()

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Walter Roberson on 8 Nov 2011
And note that any root that is output might be complex. If you want only the real roots, filter to TheRoots(imag(TheRoots)==0)
bsd on 9 Nov 2011
why is that == 0 used?
BSD
Walter Roberson on 9 Nov 2011
You can tell whether a number has a complex part or not by testing to see if the imaginary part is 0. imag(x) gives you the imaginary part of x, so imag(x)==0 tests whether the imaginary part is 0. TheRoots(imag(TheRoots)==0) thus selects only the roots which are real-valued with no imaginary component.
Of course for a quadratic function over real coefficients, either _neither_ root is complex or _both_ roots are complex...

Rick Rosson on 8 Nov 2011
x = zeros(2,1);
d = sqrt(b^2 - 4*a*c);
x(1) = ( -b + d ) / (2*a);
x(2) = ( -b - d ) / (2*a);
HTH.
Rick

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Walter Roberson on 9 Nov 2011
Beware a = 0 !
Rick Rosson on 9 Nov 2011
If a = 0, then it is not (strictly speaking) a quadratic equation. It's a linear equation, and the solution in that case is trivial to compute.
Walter Roberson on 9 Nov 2011
Yes, but it is not an uncommon problem for people to calculate or randomly generate the coefficients and forget to double-check that the system is still of the same order.