# 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.

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Um... since the double root is $-b/2a$, twice its reciprocal is $-4a/b$. And if $a=2c$ and there is a double root, then $b^2=2a^2$, which you can plug into $-4a/b$ to find the value except for its sign. Am I misunderstanding your question? –  Henning Makholm Sep 3 '11 at 23:22
The answer should be free of the $a$,$b$ or $c$.I didn't get your question either. –  Quixotic Sep 3 '11 at 23:25
Hint: The discriminant is $0$. –  André Nicolas Sep 3 '11 at 23:31

HINT $\rm\ \ \ 0\ =\ x^2 +\dfrac{b}a\ x + \dfrac{1}2\ =\ (x-r)^2\: \Rightarrow\ r^2 = \dfrac{1}2\ \Rightarrow\ \dfrac{1}r\: =\: \pm\sqrt{2}$

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+1,This would have been my approach after I realized the discriminant is zero.However one small thing which I could not understand,here we have to find $\rm\: =\: {-}\dfrac{b}c\:.$ But from $\rm\ b^2 = 8\:c^2\:,$ we would get $\pm \frac{b}{c} = \pm 2\sqrt{2}$ but why are we taking $\pm 2\sqrt{2}$ for only for $-\frac{b}{a}$? (as the answer is $\pm 2\sqrt{2}$).Thanks. –  Quixotic Sep 4 '11 at 5:54
@Foo I simplified it. –  Bill Dubuque Sep 4 '11 at 7:39

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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I think there is a small typo in the solution since $2\sqrt{2}$ is not a solution to your equation. –  user13838 Sep 4 '11 at 11:35
@percusse: $2\sqrt{2}$ is not the solution to the equation, it's the sum of the reciprocals of the two solutions. The equations have double roots at $\frac{\sqrt{2}}{2}$ or at $-\frac{\sqrt{2}}{2}$, so the sum of the reciprocals of these roots are either $2\sqrt{2}$ or $-2\sqrt{2}$. –  Arturo Magidin Sep 4 '11 at 20:46
yes, that is correct. I need more coffee :) –  user13838 Sep 4 '11 at 22:25

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 $S$.

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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, $$\frac{1}{x_1}+\frac{1}{x_2} = \frac{2}{x^*}=\pm 2{\sqrt{2}}$$

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