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).

  • 1
    $\begingroup$ 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

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 $E_0S_0 >0$,

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.