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I am interested in the zeros of this polynomial in $\lambda$: $$\lambda^2+2\lambda+1-a+\frac{ar}{\delta}=0$$ where $0<a\leq1$, $r<0$ and $\delta>0$. How to determine the sign of their real parts without explicit calculation?

Mark
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  • It'll depend on the exact values , and there are two possibilities , either both negative or one negative and one positive . And also the roots are purely real – ARROW Jul 01 '20 at 08:15
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    Hint : Vietta's theorem. But it should be mentioned that this is polynomial with respect to $\lambda$ – openspace Jul 01 '20 at 08:15
  • @openspace: thank you, I edited my post – Mark Jul 01 '20 at 08:18

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For sake of comfort, I will denote $\lambda$ by $z$. The equation is same as $$(z+1)^2=a-\frac{ar}{\delta}$$Obviously, $a-\frac{ar}{\delta}\geq 0$, and hence, both the roots are real. Now, lets write the roots precisely,$$z_1:=-1+\sqrt{a-\frac{ar}{\delta}}\quad z_2:=-1-\sqrt{a-\frac{ar}{\delta}}$$Clearly, $z_2$ is negative. However, we can't comment on the sign of $z_2$ with the provided info. For instance, when $r=-1,\delta = 3, a=0.5$ we have $z_2<0$ and when $r=-6,\delta=2,a=0.5$ we have $z_2>0$.

Anand
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