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Finding a value to get certain types of roots

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data

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

    3. The attempt at a solution

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

    The roots are real, when [tex]g \leq -1[/tex] or [tex]g \geq 3[/tex]. Negative, when [tex]g-1+ \sqrt{g^2-2g-3}<0[/tex] or [tex]g-1- \sqrt{g^2-2g-3}<0[/tex]. If [tex]g-1<- \sqrt{g^2-2g-3}[/tex] and I square both sides, then [tex]1<-3[/tex]. Here's where the troubles begin. I found out this holds true, when [tex]g \leq -1[/tex] 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. jcsd
  3. Mar 23, 2009 #2

    Dick

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    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.
     
  4. Mar 23, 2009 #3
    I followed you until here. Can you, please, explain in more detail how you arrived at this.
     
  5. Mar 23, 2009 #4

    Dick

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    I put in g=(-1) to figure out which one is larger.
     
  6. Mar 23, 2009 #5
    Thank you, Dick! Seems like I'm having one of my "D'Oh" -days again... :tongue2:
     
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