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Given the equations,

$$v_1+v_2=a-1$$

$$v_1v_2=b$$

for what ranges of $a$ and $b$, can I be sure to find $0<v_1<1$ and $0<v_2<1$. Also, for what ranges, can I be sure to find at least one $v_i$ such that $0<v_i<1$

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Really you have two equations with two variables and two parameters. The difference is that you are imagining a whole class of equations, but in any given instance the proposer will give you $a$ and $b$, then it is your job to find $v_1$ and $v_2$. –  Ross Millikan Mar 22 '11 at 4:54
    
the proposer wants to know under what $a$ and $b$ I can find solutions of $v_1$ and $v_2$ such that they satisfy the conditions. So perhaps for all $a<2b$ (obviously this is wrong, but just to give an illustration) –  picakhu Mar 22 '11 at 5:11
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I see that, but the title treated all the variables equally, while the question did not, so I thought it useful to point out the difference. Your example is in the right spirit. –  Ross Millikan Mar 22 '11 at 5:13
    
@Ross: Feel free to edit it. –  picakhu Mar 22 '11 at 5:15
    
anyway, I got $b>a-2$ and $\frac{(a-1)^{2}}{4}>b$. for the first condition. Is that right? –  picakhu Mar 22 '11 at 5:17

2 Answers 2

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The solution is $v_1=\frac{1}{2}(-1+a-\sqrt{(a-1)^2-4b}), v_2=\frac{1}{2}(-1+a+\sqrt{(a-1)^2-4b})$ or the same with $v_1, v_2$ reversed. So we need $\frac{(a-1)^2}{4} \gt b$ to keep the square roots real. Then we need $1 \lt a-\sqrt{(a-1)^2-4b}$, which with $b \gt 0$ gives $a \gt 1$. We also need $a+\sqrt{(a-1)^2-4b} \lt 3$, which leads to $b \gt a-2$. So the solution area is $a \gt 1$, $b \lt 1$ and $\frac{(a-1)^2}{4} \gt b \gt a-2$.

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I think the $(a+1)^2<2b+5$ is wrong. –  picakhu Mar 22 '11 at 16:31
    
You are right, I had some algebra wrong. Corrected. –  Ross Millikan Mar 22 '11 at 16:48
    
It appears that you also need to make sure $b>0$ is accounted for in your solution area. Should it be $\frac{(a-1)^2}{4} \gt b \gt \max(0, a-2)$ and $3 \gt a \gt 1$? –  Eric Nitardy Mar 22 '11 at 18:30
    
@Eric, yes, but I probably should have mentioned that both a and b are non-negative. –  picakhu Mar 22 '11 at 20:40

Rearranging, $v_1 + v_2 = v_1 + b/v_1 = (v_1^2 + b)/v_1 = a - 1$, so $v_1^2 + b = (a - 1)v_1$ thus $v_1^2 - (a - 1)v_1 + b = 0$. This can be solved using the quadratic formula to give $v_1 = f(a,b)$ and $v_1 = g(a,b)$, representing the two solutions to the quadratic equation, which in turn gives $v_2 = b/f(a,b)$ and $v_2 = b/g(a,b)$. Using these you should be able to answer your question.

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