# Does there exist a test, or series of tests, to ensure that a polynomial has only one real root, and it's positive?

I don't need to know solutions. I figured that the logic would be similar to computing the discriminant and testing whether it is positive. For example in quadratic systems with real coefficients,

$$ax^2 + bx + c = 0$$

if $$b^2 - 4ac < 0$$ then both solutions are complex, otherwise they are real. Then, if c > 0 they are the same sign, and then the sign of $$b$$ determines what their sign is if so.

Thus if $$b^2 - 4ac > 0$$ and $$c < 0$$ I know the quadratic equation has exactly one real, positive solution.

Are there general versions of this criteria?

• Regarding your test for quadratic systems, $-(x-1)(x-2)=-x^2+3x-2=0$ is a false positive that has two positive solutions. – peterwhy Oct 6 '18 at 20:17
• "There is only one real root, and it's positive" isn't the same thing as "there is only one positive real root". Which do you mean? I suspect this question isn't easy to answer either way... – Billy Oct 6 '18 at 20:19
• By Descartes' Rule of Signs, a polynomial has exactly one positive real root if its coefficient sequence has exactly one sign change. The converse is not true: a polynomial can have just one positive real root, while its coefficient sequence has more than one sign change. (It's required that the number of sign changes be odd, however.) – Blue Oct 6 '18 at 20:25
• @peterwhy sorry, I should have said the polynomial system is $x^2 + bx^2 + c$ (effectively dividing through by $a$). – Mike Flynn Oct 6 '18 at 20:31
• @Billy "there is only one positive real root" – Mike Flynn Oct 6 '18 at 20:34

You should look at Sturm's Theorem which tells you the number of roots a polynomial has in an interval. Taking a lower limit $$0$$ and an upper limit sufficiently large to be beyond any real zero you can check the number of positive real solutions.