Suppose [tex]p(x)=ax^2+bx+c[/tex] has two roots, reciprocals of one another. Call them [tex]r_1,r_2[/tex], with [tex]r_2=\dfrac1{r_1}[/tex].
Let's divide through [tex]p(x)[/tex] by [tex]a[/tex] for now. By the fundamental theorem of algebra, we can factorize [tex]p(x)[/tex] as
[tex]x^2+\dfrac ba x+\dfrac ca=(x-r_1)(x-r_2)=(x-r_1)\left(x-\dfrac1{r_1}\right)[/tex]
Expand the RHS to get
[tex]x^2-\left(r_1+\dfrac1{r_1}\right)x+1[/tex]
so we must have
[tex]-\dfrac ba=r_1+\dfrac1{r_1}[/tex]
[tex]\dfrac ca=1[/tex]
The first equation says [tex]a[/tex] and [tex]b[/tex] occur in a ratio of the negative sum of the roots of [tex]p(x)[/tex], while the third equation says that the first and last coefficients [tex]a,c[/tex] must be the same.
