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27 Oct 2004, 01:00
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DS. How many integers can satisfy the equation : x^2 + bx + c =0; where b, c are integers.
(1) c<0
(2) b = - (c+1)
(1) c<0
(2) b = - (c+1)
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I am not sure of this one and putting my thoughts before running to pick someone up..
quickly I think B is right.
1. does not say anything about b and hence insuff.
2. b+c = -1.
If x = 1 then b+c=-1. So we have one root of x identified clearly. Hence B is suff.
quickly I think B is right.
1. does not say anything about b and hence insuff.
2. b+c = -1.
If x = 1 then b+c=-1. So we have one root of x identified clearly. Hence B is suff.
29 Oct 2004, 02:03
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E.
1) Clearly insufficient.
2) b=-c-1 -> c=-1-b
x^2 +bx -b -1 = x^2 +b(x-1) -1 = 0.
The roos of this equation are:
x1 = 1, and x2 = b-1.
so the answer to the question: "How many roots" is 1 ot 2. Depending on the value of b. if b=2, the answer is 1.
statement 1 says that c<0. c=-1-b is negative with b=2, so we can't rule it out.
So the answer is E.
Super difficult in my opinion.
1) Clearly insufficient.
2) b=-c-1 -> c=-1-b
x^2 +bx -b -1 = x^2 +b(x-1) -1 = 0.
The roos of this equation are:
x1 = 1, and x2 = b-1.
so the answer to the question: "How many roots" is 1 ot 2. Depending on the value of b. if b=2, the answer is 1.
statement 1 says that c<0. c=-1-b is negative with b=2, so we can't rule it out.
So the answer is E.
Super difficult in my opinion.
