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?

  • $\begingroup$ Regarding your test for quadratic systems, $-(x-1)(x-2)=-x^2+3x-2=0$ is a false positive that has two positive solutions. $\endgroup$ – peterwhy Oct 6 '18 at 20:17
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    $\begingroup$ "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... $\endgroup$ – Billy Oct 6 '18 at 20:19
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    $\begingroup$ 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.) $\endgroup$ – Blue Oct 6 '18 at 20:25
  • $\begingroup$ @peterwhy sorry, I should have said the polynomial system is $x^2 + bx^2 + c$ (effectively dividing through by $a$). $\endgroup$ – Mike Flynn Oct 6 '18 at 20:31
  • $\begingroup$ @Billy "there is only one positive real root" $\endgroup$ – 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.

This is probably not quite the kind of answer you were looking for, but it does work for polynomials of arbitrary degree.


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