Given expression is 9x² + 13x +(a² + 6a) = 0
Let two reciprocal zeroes be k and 1/k.
For the general quadratic equation ax² +bx + c = 0
Product of zeroes = c/a.
Hence for given equation, product kx(1/k) = 1 = (a² + 6a) / 9
or a² + 6a - 9 = 0
from here a = [-6 + 3√13]/2, [-6 - 3√13]/2
Calculate the values again as per this method if co-efficients of a² and a are other than I have assumed