Finding a value to get certain types of roots

[Kruum]

Homework Statement

With what value of g, the roots of s^2+s(1-g)+1 are in the left half-plane (e.g. s=-2 \pm 3i) or single value in the imaginary axis (e.g. s= \pm i)

The Attempt at a Solution

s= \frac{g-1 \pm \sqrt{g^2-2g-3}}{2}. The roots are complex, if g^2-2g-3<0 \Rightarrow -1<g<3 and in the left half-plane or imaginary axis, if Re\left\{s\right\}= \frac{g-1}{2} \leq 0 \Rightarrow g=1.

The roots are real, when g \leq -1 or g \geq 3. Negative, when g-1+ \sqrt{g^2-2g-3}<0 or g-1- \sqrt{g^2-2g-3}<0. If g-1<- \sqrt{g^2-2g-3} and I square both sides, then 1<-3. Here's where the troubles begin. I found out this holds true, when g \leq -1 by slapping small and big values to my calculator, but how can I re-arrange these equations to get it.

 

[Dick]

The fact you can't solve for g by squaring mean |g-1| is never equal to sqrt(g^2-2g-3). Since you are working with g<=-1 and the sqrt is real that means one of them is always greater than the other one. In fact |g-1|>sqrt(g^2-2g-3). Another way to argue is to notice the product of the two roots is 1 (since that's the constant in the quadratic). So if both are real, they both have the same sign.

 

[Kruum]

In fact |g-1|>sqrt(g^2-2g-3).

I followed you until here. Can you, please, explain in more detail how you arrived at this.

 

[Dick]

I followed you until here. Can you, please, explain in more detail how you arrived at this.

I put in g=(-1) to figure out which one is larger.

 

[Kruum]

I put in g=(-1) to figure out which one is larger.

Thank you, Dick! Seems like I'm having one of my "D'Oh" -days again...