[Math] Finding conditions to make roots of a quadratic less than one in magnitude

Question

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.

Answer 1

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 $x_1$.

1. A first condtiion is, of course, that it has two real roots: its discriminant $\Delta$ must be positive.
2. 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: $$ap(x_0)>0, \quad ap(x_1)>0 .$$
3. 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: