$2$ equations with $4$ variables 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 $0<v_i<1$

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 $\frac{(a-1)^2}{4} \gt b \gt a-2$.  
