# Solve for inequality of AR model

I was working through my textbook and found this problem that I got stuck at: Consider the AR(2) Model $$X_t = \phi_1X_{t-1}+\phi_2X_{t-2}+\epsilon_t$$ We can assume $$\phi_2 > 0$$, so the roots of the polynomial are real numbers. Show that the inequalities for the values of $$\phi_1$$ and $$\phi_2$$ which ensure it is causal stationary are given by: $$\phi_1 + \phi_2 < 1$$ $$\phi_2 - \phi_1 < 1$$ $$|\phi_1|<1, \phi_2 < 1$$

So here's what I've done so far:

Firstly I rearranged the equation to include the backshift operator $$(1-\phi_1B-\phi_2B^2)X_t = \epsilon_t$$ Then I used the quadratic equation for the roots $$z=\frac{\phi_1±\sqrt{\phi_1^2+4\phi_2}}{-2\phi_2}$$

Except I'm not sure where to go from here. How do I calculate the inequalities that ensure it's causal stationary from here? How do I figure out the conditions?

In this framework, for causal stationarity, you need $$|z|>1$$. This means, $$|\phi_1\pm\sqrt{\phi_1^2+4\phi_2}|>2|\phi_2|=2\phi_2$$ Consider $$\phi_1>0$$ case first. Then, for the first root, we have $$\phi_1+\sqrt{\phi_1^2+4\phi_2}>2\phi_2$$, which means $$\phi_1^2+4\phi_2>4\phi_2^2+\phi_1^2-4\phi_1\phi_2 \rightarrow \phi_2-\phi_1<1$$. For the other root, we have $$-\phi_1+\sqrt{\phi_1^2+4\phi_2}>2\phi_2$$, which yields $$\phi_2+\phi_1<1$$.
For $$\phi_1<0$$, you obtain the same inequalities; and these two can be summarized as $$\phi_2+|\phi_1|<1$$, which also means $$|\phi_1|<1$$ and $$\phi_2<1$$. Don't forget that $$\phi_2$$ is positive.