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 $

The reason why the 1st method is wrong :

We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan A $ and $ \tan B $
Settings discriminant greater than equal to zero.

It seems that you meant "the sum and product of $\color{red}{\tan C}$ and $\tan B$".
Considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $0\lt B\lt \frac 34\pi$.
This means that we have to consider the condition that $f(x)=0$ has at least one solution such that $x\lt -1$ or $x\gt 0$.
Therefore, the 1st method is wrong.

The reason why the 2nd method is wrong :
In the solution in Quora,

tan B + tan C = tanB tanC - 1
tanB+tan(3pi/4 - B) = p

This step is wrong. This should be
$$\tan B+\tan\left(\frac 34\pi-B\right)=\color{red}{p-1}$$
Then, we get
$$\tan^2B-(p-1)\tan B+p=0$$
But, again, similarly as the 1st method, considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $0\lt B\lt\frac 34\pi$.
Therefore, the 2nd method is wrong.

To make the two methods correct, we have to consider the condition that
$$x^2-(p-1)x+p=0$$
has at least one solution such that $x\lt -1$ or $x\gt 0$.
Solving this gives $$p\in (-\infty,0)\cup [3+2\sqrt 2,\infty)$$
which is the same as the answer in the third method.
