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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 and $0. Also, for what ranges, can I be sure to find at least one $v_i$ such that $0

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    anyway, I got b>a-2 and \frac{(a-1)^{2}}{4}>b. for the first condition. Is that right?2011-03-22

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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    @Eric, yes, but I probably should have mentioned that both a and b are non-negative.2011-03-22
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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.