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I need to show that a function (related to enzyme kinetics) has two real, positive roots. Without giving the entire kinetics model, I have the following differential equation. By the constraints of the kinetics model this is derived from, all of the constants involved in the above equation are strictly positive.

$$\dot{C} = k_{+}(E_{0}-C)(S_{0} - C) - (k+k{-})C. $$

I have decided to define $$\frac{\dot{C}}{k_{+}} = f(c), $$ $$\frac{k+k_{-}}{k_{+}} = r, \ \mathrm{and \ thus}$$ $$f(c) = (E_{0}-C)(S_{0} - C) - rC$$

To hopefully make the problem easier to deal with. My next step was to expand the polynomial, and set it equal to 0 in order to solve for the system equilibria.

$$0 = C^2+(E_0-S_0-r)C + E_0S_0$$

Using the quadratic formula to get roots, I get that the solutions are $$\frac{-(E_0-S_0-r) \pm \sqrt{(E_0-S_0-r)^2-4E_0S_0}}{2}. $$

I need to show that both roots are real and positive. I know that to show they are real, all I need to do is show that the discriminant is positive, which I do not know how to do.

After showing that the discriminant is positive, I need to show that the whole numerator is positive to verify positivity, which I also am not sure how to do (perhaps it will become more obvious after verifying realness).

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    $\begingroup$ This is not true for all real numbers $r,S_0,E_0$ (try $r=S_0=E_0=1$) so if the statement you're trying to prove is true, there must be additional conditions you haven't told us. $\endgroup$ – saulspatz Feb 1 at 4:27
  • $\begingroup$ @saulspatz The only constraint given to me is the original quadratic equation, and that all constants are strictly positive. I will post the original quadratic so my work can be checked. $\endgroup$ – jeanquilt Feb 1 at 4:36
  • $\begingroup$ @jeanquilt the quadratic equation was mistaken, see my answer bellow. $\endgroup$ – user376343 Feb 1 at 22:41
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Presumably, those are the roots of the quadratic equation $$x^2 + (E_0 - S_0 - r)x + E_0S_0 = 0$$

As you noted, the roots are real if the discriminant is non-negative. The condition for them to be positive is that $E_0 - S_0 - r < 0$ and $E_0S_0 >0$,

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  • $\begingroup$ Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true. $\endgroup$ – jeanquilt Feb 1 at 4:48
  • $\begingroup$ If $E_0 - S_0 - r \geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero). $\endgroup$ – Michael Biro Feb 1 at 4:54
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The quadratic equation obtained from $\;f(c) = (E_{0}-C)(S_{0} - C) - rC\;$
is $$0 = C^2+(-E_0-S_0-r)C + E_0S_0.$$ The discriminant is $$(E_0+S_0+r)^2-4E_0S_0=(E_0-S_0+r)^2+4S_0r>0$$ because $S_0>0,\;r>0.$

Thus the solutions are real, equal to $$\frac{E_0+S_0+r \pm \sqrt{(E_0+S_0+r)^2-4E_0S_0}}{2},$$

and are obviously positive.

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