What do I do when there are 2 parameters? I have to find for what values of $q$ the equation $x^2-2.p.x+q-1=0$ has at least 1 root in the interval $(-2;0)$ when $p>0$.
The 2 parameters trouble me. By chance I saw that when $q=1$ the roots are $x=0$ and $x=2p$ and both aren't in the interval $(-2;0)$.
But what do I do when $q$ is NOT $=1$ ?
 A: The roots are $p+\sqrt{p^2-q+1}$ and $p-\sqrt{p^2-q+1}$. The first root is positive, so you can ignore that. So now you just have to find those $q$ for which
$$-2 < p-\sqrt{p^2-q+1} < 0$$
This gives three inequalities, all of which have to be satisfied:
$$p^2-q+1 \ge 0$$
$$p-\sqrt{p^2-q+1} > -2$$
$$p-\sqrt{p^2-q+1} < 0$$
Each of these can easily be expressed as an inequality with $q$ on the left-hand side. You will find that one of them is redundant, because it is a consequence of one of the other two.
A: $x^2+(-2p)x+(q−1)=0$
Plug this into the quadratic formula
$x = p \pm \sqrt{p^2 - (q-1)}$
since p>0 
$p  + \sqrt{p^2 - (q-1)} > 0$ and not in our interval.
$p  - \sqrt{p^2 - (q-1)} < 0\\
(q-1)<0 \\
q<1$
and 
$p  - \sqrt{p^2 - (q-1)} > -2\\
p+2 > \sqrt{p^2 - (q-1)} \\
p^2 + 4p + 4 > p^2 - q + 1\\
4p+4 > 1-q$
$-4p-3 <q < 1$
A: The discriminant is $d=p^2 - q + 1$. So a necessary condition is that $d\ge 0$, or $q\ge p^2+1$. On the other hand, $x=\pm \sqrt{d}+p$.
A: I assume that you want to find real roots of this equation.
Given $ax^2+bx+c=0$, then $D=b^2-4ac$ and if $D>0$ there exists two real roots, if $D=0$ then there is single real root, and else none. If roots exist they are given by $x_{1,2}=\frac{-b \pm \sqrt D}{2a}$.
So for $x^2-2px+q-1=0$ we have $D=4p^2-4(q-1)=4(p^2-q+1) \geq 0$ as necessary condition and since roots are given by $x_{1,2}=\frac{2p \pm 2\sqrt{p^2-q+1}}{2}$ you may compare them with given interval $(-2,0)$ and find the answer.
