Question

Find condition that zeros of polynomial p(x)=ax2+bx+c are reciprocal of each other

Answer 1

Suppose $p(x)=ax^2+bx+c$ has two roots, reciprocals of one another. Call them $r_1,r_2$, with $r_2=\dfrac1{r_1}$.

Let's divide through $p(x)$ by $a$ for now. By the fundamental theorem of algebra, we can factorize $p(x)$ as

$x^2+\dfrac ba x+\dfrac ca=(x-r_1)(x-r_2)=(x-r_1)\left(x-\dfrac1{r_1}\right)$

Expand the RHS to get

$x^2-\left(r_1+\dfrac1{r_1}\right)x+1$

so we must have

$-\dfrac ba=r_1+\dfrac1{r_1}$
$\dfrac ca=1$

The first equation says $a$ and $b$ occur in a ratio of the negative sum of the roots of $p(x)$, while the third equation says that the first and last coefficients $a,c$ must be the same.