For what possible values of $a$ does $6$ lie between the roots of the equation $x^2 + 2(a - 3)x + 9 = 0$?

Let the set of all values of $p \in R$,for which both the roots of the equation $x^2-(p+2)x+(2p+9)=0$ are negative real numbers,be the interval $(\alpha, \beta]$. Then $\beta-2\alpha$ is equal to

Match the following: Consider the equation $x^2 + 2(a - 1)x + a + 5 = 0$. Match the real values of $a$ with the conditions on the roots of the given equation.

Column-$I$	Column-$II$
$A$. Imaginary roots	$P$. $a \in (-1, 4)$
$B$. One root less than $3$ and other greater than $3$	$Q$. $a \in (-\infty, -1)$
$C$. One root less than $1$ and other greater than $3$	$R$. $a \in (-\infty, -4/3)$

The range of $a$ for which the roots of $x^2 - 2x - a^2 + 1 = 0$ lie between the roots (exclusive) of the equation $x^2 - 2(a + 1)x + a(a - 1) = 0$ is:

If $\alpha, \beta$ are the roots of $x^2+bx+c=0$,$\gamma, \delta$ are the roots of $x^2+b_1x+c_1=0$ and $\gamma < \alpha < \delta < \beta$,then $(c-c_1)^2  < $

If the minimum value of the quadratic expression $x^2+5x-2$ is $M$ and it occurs at $x=a$,then $\frac{M}{a}$ is equal to