<s>[INST] A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How could we find the sum of reciprocals of the roots of this equation? 
A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How
  could we find the sum of reciprocals of the roots of this equation?

I need some hints for solving this problem.
 [/INST] This shall answer your question.
A polynomial is of the form,
$$P(x) = (x-r_1)(x-r_2)........(x-r_n)$$
where $r_1,r_2,.....r_n $ are the roots of the polynomial.
Then the first derivative of the polynomial will be,
$$P^{'}(x) = \frac{(x-r_1)^{'}}{(x-r_1)}P(x) + \frac{(x-r_2)^{'}}{(x-r_2)}P(x) + \frac{(x-r_1)^{'}}{(x-r_1)}P(x)........\frac{(x-r_n)^{'}}{(x-r_n)}P(x)$$
since $(x-k)^{'} = 1$, we have,
$$\frac{P^{'}(x)}{P(x)} = \frac{1}{(x-r_1)} + \frac{1}{(x-r_2)} + \frac{1}{(x-r_3)} +........ \frac{1}{(x-r_n)}$$
for $x = 0$
$$\frac{P^{'}(0)}{P(0)} = \frac{1}{(-r_1)} + \frac{1}{(-r_2)} + \frac{1}{(-r_3)} +........ \frac{1}{(-r_n)}$$
$$-\frac{P^{'}(0)}{k} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} +........ \frac{1}{r_n}  $$
where $k$ is the constant term of the polynomial.
Therefore the sum of the reciprocals of the roots of any polynomial is equal to minus of the first derivative of the polynomial at $x = 0$ divided by the constant term of the polynomial.
</s>