Basic quadratic calculus Given $$ AX^2+2X-1=0 \,\ \text{where} \,\ A>0$$ 
What value of A would make the absolute value of both roots bigger than $1$?
I found the roots using quadratic formula and also found that A has to be both $[0,3]$ and $[-1,0]$ and $A>0$, does it mean such an $A$ doesn't exist? 
Thanks! 

For $A>0$ we have
\begin{align}
0&=X^2 + 2 \frac{1}{A}X - \frac{1}{A} \\
&= (X + 1/A)^2 -1/A^2 -1/A \\
&=  (X - (-1/A))^2 + (- (A+1)/A^2) \iff \\
X &= \frac{\pm\sqrt{A+1}-1}{A}
\end{align}
This is a parabola with vertex at $(-1/A, -(A+1)/A^2)$.
Then
$$
\frac{\sqrt{A+1}-1}{A}
\le \lvert X\rvert 
= \left\lvert \frac{\pm\sqrt{A+1}-1}{A} \right\rvert
\le \frac{\sqrt{A+1}+1}{A}
$$
So we need
\begin{align}
1 &< \frac{\sqrt{A+1}-1}{A} \iff \\
A + 1 &< \sqrt{A+1} \Rightarrow \\
(A + 1)^2 &< A + 1 \iff \\
A^2 + 2A + 1 &< A+1 \iff \\
A^2 + A &< 0
\end{align}
This is not possible for positive $A$, as $A^2$ is positive as well.
