How to retain only the positive root of a quadratic equation?

I was wondering if there is any Matlab function that would allow me to retain only the positive root of a quadratic equation. I want to use that function along with the 'roots' function to solve n number of quadratic equations to get n number of positive roots.

x=zeros(1,n)
for i=1:n
x(i)=function(roots([a(i) b(i) c(i)]))
end

Answer 1

[Guillaume]

To only get the positive elements of array a:

>>a = [-2.5 0 2.5 -3.5 1];
>>apos = a(a>=0)
  apos =

         0    2.5000    1.0000

[Matt J]

This has the opposite problem to Evan's answer. It doesn't handle imaginary roots

    >> a=[1+i, -1+i, 3];
    >> a(a>=0)

    ans =

       1.0000 + 1.0000i   3.0000 + 0.0000i

One should test both for positivity and realness.

[Rinu]

Got it. Thanks.

[Walter Roberson]

Joel Matthew commented on the Answer:

did you read the question ? he is asking for finding the positive root not an element in an array

[Walter Roberson]

Joel,

roots() returns a vector of values. roots() is an already known step. The question is therefore to write a function which examines a vector of values and returns the positive (real) values from the vector. So where the original poster had written

x=zeros(1,n)
for i=1:n
x(i)=function(roots([a(i) b(i) c(i)]))
end

using "function" as a stand-in for the name of the function to do the desired selection, that could be written as

x=zeros(1,n)
for i=1:n
x(i)=Extract_positive(roots([a(i) b(i) c(i)]))
end

function apos = Extract_positive(a)
  apos = a(a > 0);

which is code that only has to worry about elements of an array.

Answer 2

[Evan]

help imag

So, if you wanted to return only roots without complex parts:

R_all = roots([1 1 0 1]);
R_real = R_all(~imag(R_all))

[Rinu]

That returns both the positive and the negative root. I think your code only discards IMAGINARY roots, not NEGATIVE roots.

[Matt J]

It will discard not only imaginary roots, but any complex root with a non-zero imaginary part.

[Rinu]

Thanks.

[Evan]

Sorry! I would say I misread your question, but I originally used the word "positive" as well! My only excuse, then, is that it's a Friday. ;)