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

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Math Expert
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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
1
5
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

54% (01:37) correct 46% (01:30) wrong based on 236 sessions

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Marshall & McDonough Moderator
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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
1
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
1
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
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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 Jan 2019, 05:11
Vyshak

Can you help me explain, how x1+x2 is negative ? is it because of -b/a ? with b as positive and a =1 ?


Vyshak wrote:
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 09 Jan 2019, 17:33
1
Shrey9 wrote:
Vyshak

Can you help me explain, how x1+x2 is negative ? is it because of -b/a ? with b as positive and a =1 ?


Vyshak wrote:
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


Hi Shrey9

yes you are correct. Sum of roots is -b/a so if b is positive and in this equation a=1, -b has to be negative
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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 Jan 2019, 21:36
1
Bunuel wrote:
For integers x, b and c, x^2 + bx + c = 0. Is x > 0?

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


Comparing \(x^2\) + bx + c =0 with the general form of Quadratic Eqn \(ax^2 + bx + c = 0\)

We get a =1, b = b , and c = c

Let \(x_1\) and \(x_2\) be the two roots of the equation

Sum of the roots = \(x_1 + x_2\) =\(\frac{-b}{a}\) = \(\frac{-b}{1}\)= -b

Product of the roots = \(x_1 * x_2\) = \(\frac{c}{a}\) = \(\frac{c}{1}\) = c

St 1 :

b > 0, i.e -b < 0

\(x_1 + x_2 < 0\)

Let us plug in some values

Case 1 : \(x_1 = - 3, x_2 = - 2\) Both negative values

Case 2 : \(x_1 = - 3, x_2 = 2\) One value positive, one value negative

Not Sufficient

St 2 :

c > 0

\(x_1 * x_2 > 0\)

\(x_1 and x_2\) have the same sign

Both positive or Both negative

Not Sufficient

St 1 and St 2 Combined

Only Both negative remains

Sufficient
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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 21 Jan 2019, 08:23
Top Contributor
Bunuel wrote:
For integers x, b and c, x² + bx + c = 0. Is x > 0?

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


Target question: Is x > 0?

Statement 1: b > 0
Let's TEST some cases that satisfy statement 1:
If b = 5, and c = -14, the equation becomes x² + 5x - 14 = 0. Factor to get: (x + 7)(x - 2), which means EITHER x = -7 OR x = 2
If x = -7, then the answer to the target question is NO, x is NOT greater than 0
If x = 2, then the answer to the target question is YES, x is greater than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: c > 0
Let's TEST some cases that satisfy statement 2:
Case a: b = -6, and c = 9. The equation becomes x² - 6x + 9 = 0. Factor to get: (x - 3)(x - 3), which means x = 3. So, the answer to the target question is YES, x is greater than 0
Case b: b = 6, and c = 9. The equation becomes x² + 6x + 9 = 0. Factor to get: (x + 3)(x + 3), which means x = -3. So, the answer to the target question is NO, x is NOT greater than 0

Statements 1 and 2 combined
We have: x² + bx + c = 0
Let's say that, when we factor x² + bx + c we get: x² + bx + c = (x + j)(x + k)
If c is POSITIVE, we know that EITHER j and k are both positive, OR j and k are both negative

We also know that, since b is POSITIVE, the sum of j and k must be POSITIVE

This allows us to rule out the possibility that j and k are both negative, since two negative numbers cannot add up to be positive.

So, it MUST be the case that j and k are both positive
In other words, x² + bx + c = (x + positive)(x + positive)
If (x + positive)(x + positive) = 0, then we can see that two solutions will both be negative.
For example, if (x + 2)(x + 3) = 0, then the two solutions are x = -2 and x = -3
Similarly, if (x + 1)(x + 10) = 0, then the two solutions are x = -1 and x = -10
If the two solutions will both be NEGATIVE, the answer to the target question is NO, x is NOT greater than 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0?   [#permalink] 21 Jan 2019, 08:23
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