Can an unstable plant be stabilized with a lead-lag controller?

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I have the following plant transfer function

$$P(s) = 500\cdot\frac{s-5}{s^2 - 625} $$

this function is not stable. I want to know if the system can be stable with a controller of the form

$$ C(s) = k\cdot\frac{s+a}{s+b}$$

I can see mathematically how it's not possible, since (non-unity) closed loop transfer function: $$ \frac{1}{1 + P(s) \cdot C(s)}$$ is not stable, but I wanted to know if there is some other way of knowing that a lead-lag controller will not stabilize this system, maybe a deeper insight to what a lead-lag can or cannot do?

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As KBS suggested, we can look at the system's root locus to gain a deeper insight into what's happening.

System's root locus

That pole-zero pair on the right side of the $s$-plane is a problem. There's no amount of poles or zeros that we can add to the left side that will bring all closed-loop poles there for any gain value.

Our only hope to stabilise this system with simple poles and zeros is to add a pole in between that right side pole-zero pair, so as to create a breakaway point from the real axis, and a suitable zero in the left side that will allow us to move the closed-loop poles to the left side for some gain value.

Compensated root locus

For completeness, a causal second-order system of the form $$ K{s+a \over (s-b)(s+c)} $$ can stabilize the plant, but then the compensator itself is open-loop unstable.

Root locus, 2nd-order compensator