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Math Expert V
Joined: 02 Sep 2009
Posts: 57155
For integers x, b and c, x^2 + bx + c = 0. Is x > 0?  [#permalink]

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5 00:00

Difficulty:   55% (hard)

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

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

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

_________________
Marshall & McDonough Moderator D
Joined: 13 Apr 2015
Posts: 1680
Location: India
Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0?  [#permalink]

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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
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1193
Location: India
GPA: 3.82
Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0?  [#permalink]

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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
Manager  S
Joined: 23 Apr 2018
Posts: 112
Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0?  [#permalink]

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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
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1193
Location: India
GPA: 3.82
Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0?  [#permalink]

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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
GMAT Tutor B
Joined: 07 Nov 2016
Posts: 56
GMAT 1: 760 Q51 V42 For integers x, b and c, x^2 + bx + c = 0. Is x > 0?  [#permalink]

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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
_________________
Please contact me at ksrutisagar@gmail.com
CEO  V
Joined: 12 Sep 2015
Posts: 3912
Location: Canada
Re: For integers x, b and c, x^2 + bx + c = 0. Is x > 0?  [#permalink]

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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
_________________ 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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