Finding conditions to make roots of a quadratic less than one in magnitude I'm doing a problem that asks for you to find the conditions that make $y$ defined:
$$y=x^2-bx+c$$
have real roots with magnitude less than one.
Now the condition for the roots being real seems to be: $$b^2\ge4c$$ The problem I have is finding the restrictions necessary for the second condition to be true, since intuitively it seems it should be that:
$$ \left|\frac{b\pm\sqrt{b^2-4c}}{2}\right|<1 \;\;\rightarrow\;\;\left|\,b\pm\sqrt{b^2-4c}\right|<2 $$
I'm not sure how to tackle the problem from here effectively (or indeed if this is the best way to tackle this type of question) since when I try to evaluate cases of the absolute value (using its definition) they seem to give contradictory results, for example I can't see where conditions that $|\,b\,|<2$ come from. From playing with Mathematica the answer it gives is:
$$ (-2 < b \leq 0\; \;\land\;\; -b - 1 < c \leq \frac{b^2}{4}) \;\;\lor \;\; (0 < b < 2 \;\;\land 
   \;\;b - 1 < c \leq \frac{b^2}{4}) $$
Which seems to make sense, at least trying values in those regions seem to work. I'm just wondering the best technique to tackle this kind of problem. 
 A: A quick thought:
Let $y=(x-α)(x-β)$
Our roots are $x=α$ and $x=β$, and we want their magnitudes to be less than 1.
We also have $b=α+β$ and $c=αβ$.
If $|\,b\,|≥2$, we have $2≤|\,b\,|=|\,α+β\,|≤|\,α\,|+|\,β\,|$ so either $|\,α\,|≥1$ or $|\,β\,|≥1$.
For the condition on $c$, one comes from the discriminant as you mentioned, and you can get the other by looking at the graph of $y$ at $x=1$ and $x=-1$.
A: There is a generic way to solve this classical problem:
Find conditions on the coefficients of a quadratic polynomial $p(x)=ax^2+
bx+c$ so that it has two real roots between $x_0$  and $x_1$.


*

*A first condtiion is, of course, that it has two real roots: its discriminant $\Delta$ must be positive.

*Let $\alpha<\beta\,$ be the real roots. A second condition is  that $x_0$ and $x_1$  must be outside the interval $[x_0, x_1]$. This means:
$$ap(x_0)>0, \quad ap(x_1)>0 .$$

*The last condition is that $\,x_0<\alpha$ and $\beta>x_1$. Knowing 2, this is equivalent to $$x_0<\dfrac{\alpha+\beta}2<x_1\iff x_0<-\frac b{2a}<x_1$$
In the present case, all this translates to:
$$  \begin{cases}
b^2>4c\\
b+c>-1,\quad b-c<1\\-2<b<2
\end{cases} $$
Graphical representation of the solutions as a domain of the plane:

