# Finding a value to get certain types of roots

1. Mar 23, 2009

### Kruum

1. The problem statement, all variables and given/known data

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. $$s= \pm i$$)

3. 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.

Last edited: Mar 23, 2009
2. Mar 23, 2009

### 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.

3. Mar 23, 2009

### Kruum

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

4. Mar 23, 2009

### Dick

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

5. Mar 23, 2009

### Kruum

Thank you, Dick! Seems like I'm having one of my "D'Oh" -days again... :tongue2: