Question

Please help me with this question Number 13

Answer 1

hey listen
its a quadratic equation and we know that the product of the zeros is
C/A and here in the question its given that the roots are resiprocal to each other so their product will be 1.
6a/$a^{2}$+9=1
6a=$a^{2}$+9
$a^{2}$-6a+9=0
$(a-3)^{2}$=0
a=+3/-3