Proving 0 < θ1 < β in θ12-γθ1+β=0

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Kinetica
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Homework Statement



θ12-γθ1+β=0

Show that

0<θ1


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.
 
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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.

[tex]\theta_{1} = \frac{\gamma\pm\sqrt{\gamma^{2}-4\beta}}{2}[/tex]

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?
 
consider for instance β=0 and y=θ then θ can be whatever you like as long as y is also!
 
OK so I found this on your other request, and it looks like it could apply here.
Kinetica said:
θ12
θ12
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...
 
Kinetica said:

Homework Statement



θ12-γθ1+β=0

Show that

0<θ1

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.
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 .