ensure stability against coefficient quantization
your filter is in the form of a standard biquad with 5 independent coefficients ($b_2=0$ in your case):

$$ H(z) = \frac{b_0 + b_1 z^{-1} + b_2 z^{-2}}{ 1 + a_1 z^{-1} + a_2 z^{-2}} $$

your two poles are the value of $z$ that make the denominator equal to zero

$$ 1 + a_1 z_p^{-1} + a_2 z_p^{-2} = 0 $$

or (multiplying both sides by $z_p^2$),

$$ z_p^2 + a_1 z_p + a_2 = 0 \tag{1} $$

and stability depends entirely on whether your poles (after coefficient quantization) are contained in the unit circle:

$$ |z_p| &lt; 1 $$

or

$$ |z_p|^2 &lt; 1 $$

the solution to Eq. 1 is (quadratic formula):

$$ z_p = -\frac{a_1}{2} \pm \sqrt{\left(\frac{a_1}{2}\right)^2 - a_2} $$

and, if the poles are complex conjugate, then $\left(\frac{a_1}{2}\right)^2 &lt; a_2$ and your complex conjugate poles are

$$ z_p = -\frac{a_1}{2} \pm j \sqrt{a_2 - \left(\frac{a_1}{2}\right)^2} $$

in the case of two real poles, when $\left(\frac{a_1}{2}\right)^2 \ge a_2$, you must guarantee (to ensure stability):

$$ -1 &lt; -\frac{a_1}{2} \pm \sqrt{\left(\frac{a_1}{2}\right)^2 - a_2} &lt; +1 $$

that means both of the following must be guaranteed (after quantization):

$$ -\frac{a_1}{2} + \sqrt{\left(\frac{a_1}{2}\right)^2 - a_2} &lt; +1 $$

$$ -1 &lt; -\frac{a_1}{2} - \sqrt{\left(\frac{a_1}{2}\right)^2 - a_2} $$

what is necessary (but not sufficient) is that $|a_1| &lt; 2$ and what also must be true is (getting rid of the $\sqrt{\cdot}$)

$$ \left(\frac{a_1}{2}\right)^2 - a_2 &lt; \left(1 + \frac{a_1}{2}\right)^2 = 1 + a_1 + \left(\frac{a_1}{2}\right)^2 $$

and

$$ \left(\frac{a_1}{2}\right)^2 - a_2 &lt; \left(1 - \frac{a_1}{2}\right)^2  = 1 - a_1 + \left(\frac{a_1}{2}\right)^2 $$

or 

$$ a_2 &gt; -1 - a_1 $$

$$ a_2 &gt; -1 + a_1 $$

so for real poles ($\left(\frac{a_1}{2}\right)^2 \ge a_2$) you must make sure, after quantization that

$$|a_1| &lt; 2$$
and
$$-1 + |a_1| &lt; a_2 \le \left(\frac{a_1}{2}\right)^2$$

for complex conjugate poles ($\left(\frac{a_1}{2}\right)^2 &lt; a_2$), it's much, much easier.  i'll leave this for an exercise for the reader but you can easily show that (for complex conjugate):

$$ |z_p|^2 = a_2 $$

which means

$$ 0 &lt; |z_p|^2 = a_2 &lt; 1 $$

and you still will see that $|a_1| &lt; 2$ is necessary because $|a_1| &lt; 2 a_2 &lt; 2$.  it turns out that $-\frac{a_1}{2}$ is the point on the $\mathfrak{Re}(z)$ axis that is midway between the two poles, whether they be real or complex conjugate.

now, if your digital filter is slewing from one stable complex pole case to another stable complex pole place, then $a_2$ is slewing from one value less than 1 to another value less than one and all filters in between will have $a_2$ less than one and will be stable, at least in a "static" sense.  rapid coefficient slewing can create instabilities in and of itself, but if the slewing stops (at a stable condition), any instability also stops.

so in both real and complex conjugate cases, to have stability in the above biquad you must have

$$ -2 &lt; a_1 &lt; 2 $$

and

$$ 0 &lt; a_2 &lt; 1 $$

for the complex conjugate case $a_1^2 &lt; 4 a_2$

or

$$-1 + |a_1| &lt; a_2 \le \left(\frac{a_1}{2}\right)^2 = \frac{|a_1|^2}{4}$$

for the real pair case $a_1^2 \ge 4 a_2$.

note that, in the real pole pair case, we know that $$-1 + |a_1| &lt;  \frac{|a_1|^2}{4}$$ because $$-1 + |a_1| - \frac{|a_1|^2}{4} = -\frac{1}{4}(|a_1|-2)^2 &lt; 0$$.  so there is always a gap for $a_2$ to fit into.