<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] 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 $-2\sqrt{2}$. 
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