# Show theta<beta

1. Jan 19, 2012

### Kinetica

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

θ12-γθ1+β=0

Show that

0<θ1

3. The attempt at a solution

I know that for θ1=0, θ12-γθ1+β>0:
Substituting, we get 02-0+β=β, which is positive.

I don't know how to show that for θ1=λ, θ12-γθ1+β<0.

I also don't know how to show that these results imply that there is zero between these two values. Which in turns means that 0<θ1<β.

2. Jan 19, 2012

### Joffan

Well, θ12-γθ1+β=0 is just a quadratic equation in θ1, so there should be at most two possible values for θ1, in terms of the other parameters.

$$\theta_{1} = \frac{\gamma\pm\sqrt{\gamma^{2}-4\beta}}{2}$$

But as things stand, you haven't given nearly enough information to assert the inequalities required. For example, why shouldn't β be zero or even negative? Is γ greater than zero, greater than one?

3. Jan 19, 2012

### conquest

consider for instance β=0 and y=θ then θ can be whatever you like as long as y is also!

4. Jan 19, 2012

### Joffan

OK so I found this on your other request, and it looks like it could apply here.
but that just means that the relative sizes of θ1 and β depend on θ2. So still no closer to that inequality - there's something you aren't telling us about this question...

5. Jan 19, 2012

### SammyS

Staff Emeritus
This thread is very similar in topic to a thread you started one day earlier: https://www.physicsforums.com/showthread.php?p=3717073#post3717073 .

Where did λ (lambda) come from, or should that be γ (gamma) ?

I suppose we can infer that β > 0 from the inequality, 0<θ1<β, and because you mentioned it in passing, "Substituting, we get 02-0+β=β, which is positive."

It makes no sense to plug values such as 0 or γ or λ in for θ1 in the quadratic polynomial θ12-γθ1+β to see if 0<θ1<β .

As Joffan said, you need more information regarding β and γ, before you can say much about θ1 .