When roots of $х^2 + рх + q = 0$ are reciprocal ($a$ and $1/a$)?

Question

1) When roots of $х^2 + рх + q = 0$ are reciprocal and same sign ($a$ and $1/a$) ?

2) When roots of $х^2 + рх + q = 0$ are reciprocal and different sign ($-a$ and $1/a$) ?

I see that for (1) $q=1$ (q shall be only 1), for (2) $q=-1$ but what limitations shall I attach to $p$?

I understand that $x_1 + x_2 = -p$, but it is not enough.

For example $х^2 - х - 1 = 0$ is a brilliant one of the family of (2)-type equations (with reciprocal roots of different signs): $-\phi$ & $1/\phi$ (golden ratio).

P.S. Assumed equation has two different roots.

Answer 1

It's pretty straight forward that is the roots of such a polynomial equation are a and 1/a we have (x- a)(x- 1/a)= x^2- ax- x/a+ 1= x^2- (a+ 1/a)x+ 1 which is x^2+ px+ q if and only if q= 1 and p= -(a+ 1/a).

The roots are a and -1/a if (x-a)(x+1/a)= x^2+ (1/a- a)x- 1. q= -1 and p= 1/a- a.

Answer 2

By Vieta's formulas the product of the roots of $x^2+px+q$ is $q$, but such roots are real iff $p^2\geq 4q$.

So any polynomial of the $x^2+px-1$ kind has real real roots of the $-a,1/a$ kind, while the polynomials $x^2+px+1$ have real roots of the $a,1/a$ kind only if $|p|\geq 2$.