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.
 A: HINT $\rm\ \ \ 0\ =\ x^2 +\dfrac{b}a\ x + \dfrac{1}2\ =\ (x-r)^2\: \Rightarrow\ r^2 = \dfrac{1}2\ \Rightarrow\ \dfrac{1}r\: =\: \pm\sqrt{2}$
A: Since the roots are equal, you must have $b^2 = 4ac = 8c^2$. Also, the roots satisfy
$$0=x^2 + \frac{b}{a}x + \frac{c}{a} = x^2 + \frac{b}{a} + \frac{1}{2},$$
so the product of the roots is $\frac{1}{2}$; that is, $4c^2 = \frac{1}{2}$. Therefore, $b^2=2(4c^2) = 1$. 
The sum of the roots is $-\frac{b}{a}$; but it also is equal to $2a$. So $2a = -\frac{b}{a}$. This gives $2a^2 = -b$, and we know $|b|=1$, so we must have $b=-1$. So the sum of the reciprocals is $-4a/b = 4a$. 
Now, $a^2 = \frac{1}{2}$, so either $a=\frac{\sqrt{2}}{2}$ or $a=-\frac{\sqrt{2}}{2}$. Thus, the equation is either
$$\frac{\sqrt{2}}{2}x^2 - x + \frac{\sqrt{2}}{4}\quad\text{or}\quad -\frac{\sqrt{2}}{2}x^2 - x -\frac{\sqrt{2}}{4}.$$
Both satisfy the desired conditions: they have a double root at $a=2c$. In one case, the sum of the reciprocals is $2\sqrt{2}$, in the other it is $-2\sqrt{2}$. 
A: If neither of the roots $x_1,x_2$ of a quadratic equation is $0$, then the sum of the reciprocals of the roots is
$$S=\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}.$$
Note that the numerator is the sum and the denominator is the product of the roots. What are these quantities in terms of the coefficients of the equation? Substitute and you get a representation of $S$ in terms of two of the coefficients.
At this point, to determine $S$ consider your specific equation with the conditions of equal roots (value of discriminant?) and $a=2c$. With a little bit of equation manipulation you get two possible values for $S$.
A: Also, just for the fun of it, since you have a root with multiplicity two:
$$
ax^2+bx+c = (\sqrt{a}x+\sqrt{c})^2
$$
Then the root is $x^* = \pm\sqrt{\frac{c}{a}}$ and from $a = 2c$, we get $x^* = \pm\sqrt{\frac{1}{2}}$. Hence, $$
\frac{1}{x_1}+\frac{1}{x_2} = \frac{2}{x^*}=\pm 2{\sqrt{2}}$$
A: 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.
