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# What is the sum of the reciprocals of the roots of the equation

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Joined: 02 Sep 2009
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What is the sum of the reciprocals of the roots of the equation  [#permalink]

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19 Mar 2019, 21:30
00:00

Difficulty:

65% (hard)

Question Stats:

35% (02:01) correct 65% (02:19) wrong based on 20 sessions

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What is the sum of the reciprocals of the roots of the equation $$\frac{2003}{2004}*x+1+\frac{1}{x}=0$$?

(A) -2004/2003
(B) -1
(C) 2003/2004
(D) 1
(E) 2004/2003

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What is the sum of the reciprocals of the roots of the equation  [#permalink]

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19 Mar 2019, 22:41
Bunuel wrote:
What is the sum of the reciprocals of the roots of the equation $$\frac{2003}{2004}*x+1+\frac{1}{x}=0$$?

(A) -2004/2003
(B) -1
(C) 2003/2004
(D) 1
(E) 2004/2003

Simplifying the equation

$$\frac{2003}{2004}*x+1+\frac{1}{x}=0$$

$$2003x^2 + 2004x +2004 = 0$$

For any equation $$ax^2 + bx + c = 0$$
Smm of the roots (p and q), $$p+q = \frac{-b}{a} = \frac{-2004}{2003}$$
Product of the roots, $$p*q = \frac{c}{a} = \frac{2004}{2003}$$

Required information $$= \frac{1}{p} + \frac{1}{q} =\frac{(p+q)}{pq} = (-2004/2003) / (2004/2003) = -1$$

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Re: What is the sum of the reciprocals of the roots of the equation  [#permalink]

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20 Mar 2019, 00:20
Bunuel wrote:
What is the sum of the reciprocals of the roots of the equation $$\frac{2003}{2004}*x+1+\frac{1}{x}=0$$?

(A) -2004/2003
(B) -1
(C) 2003/2004
(D) 1
(E) 2004/2003

$$\frac{2003}{2004}*x+1+\frac{1}{x}=0$$
2003x^2+2004x+2004=0
a=2003
b=2004
c=2004
sum = -b/a = -2004/2003 = a+b
product = c/a = 2004/2003; a*b
so
1/a+1/b = a+b/ab ; -2004/2003 * 2003/2004 = -1
IMO B
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Re: What is the sum of the reciprocals of the roots of the equation  [#permalink]

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20 Mar 2019, 20:34
2003/2004∗x^2+x+1 =0
b^2 - 4 a c = 1-4(2003/2004)(1) and it's <0 , so the equation has no roots
please correct me if I'm wrong
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Re: What is the sum of the reciprocals of the roots of the equation   [#permalink] 20 Mar 2019, 20:34
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