Closed-loop pole moving to RHP - less stable or more stable?

Question

I was reading Razavi's Design of Analog CMOS ICs book and came across this in the Stability and Freq. Compensation chapter,
What the book says
The closed-loop transfer function of a system is given by the following:

It then goes on to say that,

Okay, that makes sense to me. As loop-gain (beta or Aol) increases, our pole moves further into the LHP, making any oscillation die out much quicker. Now, a few pages later, the book has this,

He says here that by reducing the feedback factor (hence loop gain) our system is more stable and this makes sense to me from the bode plot.
My Question
The two bolded statements above contradict each other. If I reduce my feedback factor (hence loop gain), my pole will move towards the RHP and thus from the s-plane perspective, the system is less stable but from the bode plot perspective, the system is more stable. Why does the s-plane and bode plot contradict each other?

rpm2718 · Answer

You are comparing a first-order system with a second-order system.
The first equation you show (10.4) describes a first-order system. The pole moves farther along the negative real axis as you increase the feedback, as the text describes.
The Bode plot in and discussion around Figure 10.7 is a second-order system. In a second order system with real open-loop poles, the poles first move toward one another along the real axis with increasing feedback, then after meeting along the real axis, move away from one another in the positive and negative imaginary directions. Thus the angle they make with the imaginary axis is decreasing, leading to a less stable system with increasing feedback. Thus the Bode analysis and the root-locus are consistent.

Closed-loop pole moving to RHP - less stable or more stable?

What the book says

My Question

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