Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and  $C$  are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives  $ p \le 3 - 2\sqrt2 $. Or  $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B  = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know   $ \cos (B+C)  =  1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$.  $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives 
$ P \lt 0 $  or $ p \ge 3+ 2\sqrt2 $

We should consider discriminants  but in the way they arise  directly:
Let $tx = \tan(x)$ for shorthand notation
$$t_A= t{(\pi-B-C)}$$
$$1=\frac{-(tA+tB)}{1-tA\, tB} $$
$$tA \,tB=p\quad$$
plug in and simplify to find quadratic equation roots
$$ tB^2+(1-p)tB+p=0 $$
$$ 2\,tA=-(1-p)-\sqrt{1-6p+p^2} $$
$$ 2\,tB=-(1-p)+\sqrt{1-6p+p^2}$$
For this to be real, quantity under radical sign should  be positive. 
The roots of quadratic equation then supply upper and lower limits automatically.
$$p_{lower}= 3+2\sqrt{2}\quad p_{lower}= 3-2\sqrt{2} $$
when $p$ should not lie between these limits as the quantity under radical sign  is: 
$$ \sqrt{p^2 -6p +1 } = \sqrt{(p-p_{lower})(p-p_{upper})} $$
