How many integers can let x^2+bx+c=0? B and c are constants.

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How many integers can let x^2+bx+c=0? B and c are constants. [#permalink]

18 Aug 2006, 23:33

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How many integers can let x^2+bx+c=0? B and c are constants.
1) c<0
2) b= - (c+1)

Hi guys shouldnt the answer to this question be c? Accdg to the question bank its E.

b^2-4ac--> c^2+1-2c which given 1 and 2 would always be greater than zero

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19 Aug 2006, 02:27

E is correct.

roots are [-b + SQRT(b^2 - 4c)]/2
and [-b + SQRT(b^2 - 4c)]/2

St1: b^2 - 4c is always +ve. There are two real roots but we can not say those will be integers or not: INSUFF

St2: b^2 - 4c = (c+1)^2 - 4c = (c-1)^2: There are two real roots but we can not say those will be integers or not: INSUFF

Together:
Same as above.: INSUFF
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19 Aug 2006, 13:00

From stmt 2,
b = -(c+1)
=> c = -b-1
Substituting this value in the formula
b^2-4c
= b^2 - 4(-b-1)
= b^2+4b+4
= (b+2)^2
Substituting in the formula for roots of quadratic eqn
[-b +or- (b+2)]/2
=1 or
= -b-1
Thus, we will always get 2 integral roots.
Hence soln. should be (B).
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Prashrash.

[#permalink] 19 Aug 2006, 13:00

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How many integers can let x^2+bx+c=0? B and c are constants.

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How many integers can let x^2+bx+c=0? B and c are constants.

How many integers can let x^2+bx+c=0? B and c are constants.

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