Question 966986
If the roots of {{{5x^2+13x+p=0}}} are {{{r}}} and {{{1/r}}} ,
then we can "factorize" {{{5x^2+13x+p}}} like this
{{{5x^2+13x+p=5(x-r)(x-1/r)}}} ,
and if we multiply the two expressions in brackets, we get
{{{5x^2+13x+p=5(x-r)(x-1/r)=5(x^2-(1/r)x-rx+r(1/r))=5(x^2-(1/r+r)x+1)=5x^2-5(1/r+r)x+5}}} .
So, since that is true for all values of x,
the coefficients must be the equal, meaning that
{{{highlight(p=5)}}} ,
and {{{13=-5(1/r+r)}}} .
We can find those reciprocal roots,
or at least prove that there are real roots,
to make sure we have not been tricked.
It turns out that the determinant for {{{5x^2+13x+5=0}}} is
{{{13^3-4*5*5=169-100>0}}} and there are real roots given by
{{{x=(-13 +- sqrt(69))/10}}} .