A quadratic equation ax2+bx+c=0 has equal roots at a=2c How could we find the sum...

Step 1
Since the roots are equal, you must have
${\displaystyle b^2=4ac=8c^2}$
Also, the roots satisfy
${\displaystyle 0=x^2+\frac{b}{a}x+\frac{c}{a}}$
${\displaystyle =x^2+\frac{b}{a}+\frac{1}{2}}$
so the product of the roots is ${\displaystyle \frac{1}{2}}$; that is, ${\displaystyle 4c^2=\frac{1}{2}.}$ Therefore, ${\displaystyle b^2=2\left(4c^2\right)=1}$
The sum of the roots is ${\displaystyle -\frac{b}{a}}$; but it also is equal to 2a. So
${\displaystyle 2a=-\frac{b}{a}}$
This gives
${\displaystyle 2a^2=-b}$,
and we know
${\displaystyle \left|b\right|=1}$,
so we must have
${\displaystyle b=-1}$
So the sum of the reciprocals is
${\displaystyle -\frac{4a}{b}=4a}$
Now, ${\displaystyle a^2=\frac{1}{2}}$
so either
${\displaystyle a=\frac{\sqrt{2}}{2}}$
or
${\displaystyle a=-\frac{\sqrt{2}}{2}}$
Thus, the equation is either
${\displaystyle \frac{\sqrt{2}}{2}{x}^2-x+\frac{\sqrt{2}}{4}}$
or
${\displaystyle -\frac{\sqrt{2}}{2}{x}^2-x-\frac{\sqrt{2}}{4}}$
Both satisfy the desired conditions: they have a double root at ${\displaystyle a=2c}$. In one case, the sum of the reciprocals is ${\displaystyle 2\sqrt{2}}$, in the other it is ${\displaystyle -2\sqrt{2}}$.

Step 1
Also, just for the fun of it, since you have a root with multiplicity two:
${\displaystyle a{x}^2+bx+c={(\sqrt{ax}+\sqrt{c})}^2}$
Then the root is
${\displaystyle x^{\cdot }=\pm \sqrt{\frac{c}{a}}}$
and from ${\displaystyle a=2c}$, we get
${\displaystyle x^{\cdot }=\pm \sqrt{\frac{1}{2}}}$
Hence,
${\displaystyle \frac{1}{x_1}+\frac{1}{x_2}=\frac{2}{x^{\cdot }}=\pm 2\sqrt{2}}$

Step 1This shall answer your question. A polynomial is of the form,$P(x)=(x-r_1)(x-r_2)\cdots (x-r_n)$where $r_1,r_2,\cdots r_n$ are the roots of the polynomial. Then the first derivative of the polynomial will be,$P^{\prime }(x)=\frac{(x-r_1{)}^{\prime }}{(x-r_1)}P(x)+\frac{(x-r_2{)}^{\prime }}{(x-r_2)}P(x)+\frac{(x-r_1{)}^{\prime }}{(x-r_1)}P(x)\cdots \frac{(x-r-n{)}^{\prime }}{(x-r_n)}P(x)$since $(x-k{)}^{\prime }=1$, we have,$\frac{P^{\prime }(x)}{P(x)}=\frac{1}{(x-r_1)}+\frac{1}{(x-r_2)}+\frac{1}{(x-r_3)}+\cdots \frac{1}{(x-r_n)}$for $x=0$$\frac{P^{\prime }(0)}{P(0)}=\frac{1}{(-r_1)}+\frac{1}{(-r_2)}+\frac{1}{(-r_3)}+\cdots \frac{1}{(-r_n)}$$-\frac{P^{\prime }(0)}{k}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\cdots \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.