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For integers x, b and c, x^2 + bx + c = 0. Is x > 0?

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Joined: 02 Sep 2009
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For integers x, b and c, x^2 + bx + c = 0. Is x > 0? [#permalink]

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New post 09 Feb 2017, 01:23
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D
E

Difficulty:

  55% (hard)

Question Stats:

59% (01:09) correct 41% (01:08) wrong based on 112 sessions

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Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0? [#permalink]

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New post 09 Feb 2017, 09:42
x^2 + bx + c = 0 can have two roots x1 and x2

Sum of the roots = x1 + x2 = -b
Product of the roots =x1 * x2 = c

St1: b > 0 --> x1 + x2 is negative. Cannot determine if the roots are positive or negative.
Not Sufficient.

St2: c > 0 --> x1 * x2 is positive --> Both the roots are either positive or negative.
Not Sufficient.

Combining St1 and St2,
x1 + x2 = negative and x1 * x2 = positive --> Both the roots have to be negative.
Sufficient.

Answer: C
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Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0? [#permalink]

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New post 15 Feb 2018, 10:51
Bunuel wrote:
For integers x, b and c, x^2 + bx + c = 0. Is x > 0?

(1) b > 0
(2) c > 0


One of the method is to use the properties of roots of a quadratic equation that has already been explained in the earlier post. We can also solve this question logically.

as \(x^2 + bx + c = 0\) and \(x^2\) is always positive so the summation to be \(0\) either both \(bx+c\) is negative with magnitude equal to \(x^2\) or either of \(bx\) or \(c\) is negative.

Statement 1: given \(b>0\), if \(c>0\) then \(x<0\) but if \(c<0\) then \(x>0\) or \(x<0\). Hence multiple scenarios possible. Insufficient

Statement 2: given \(c>0\), if \(b>0\) then \(x<0\) and if \(b<0\) then \(x>0\). Again multiple scenarios possible. Insufficient

Combining 1 & 2: \(b>0\) & \(c>0\), hence \(x<0\) . Sufficient

Option C
Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0?   [#permalink] 15 Feb 2018, 10:51
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For integers x, b and c, x^2 + bx + c = 0. Is x > 0?

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