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Manager  Joined: 10 Feb 2011
Posts: 102
If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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4
4 00:00

Difficulty:   45% (medium)

Question Stats: 65% (01:36) correct 35% (01:38) wrong based on 172 sessions

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If x^2 + 12x + 20 < 0, how many integer values can x be?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Manager  Joined: 24 Nov 2010
Posts: 152
Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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banksy wrote:
27. (SC) If x^2 + 12x + 20 < 0, how many integer values can x be?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

x^2 + 12x + 20 < 0 --> (x+10)(x+2)<0 this will be true only when x<-2 and x> -10.

Ans = C
Manager  Joined: 18 Oct 2010
Posts: 57
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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the same solution and also found -8<a<-2 and
Retired Moderator Joined: 20 Dec 2010
Posts: 1463
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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1
1
banksy wrote:
27. (SC) If x^2 + 12x + 20 < 0, how many integer values can x be?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Sol:

x^2+12x+20 = x^2+10x+2x+20 = x(x+10)+2(x+10) = (x+2)(x+10)

(x+2)(x+10) < 0

Roots of x= -10,-2

Ranges; x<-10, -10<x<-2, x>-2

For x>-2, let's say x=0; (x+2)(x+10) = 2*10=20. It is a positive

Thus,
x>-2 is +ve
-10<x<-2 is -ve; alternated from previous
x<-10 is +ve; alternated from previous

For "<" sign, we must consider only -ves.

-10<x<-2 befits the value of x that satisfy the inequality.
x can be -9,-8,-7,-6,-5,-4,-3

Count=7

Ans: "B"
Senior Manager  S
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 382
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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fluke wrote:
banksy wrote:
27. (SC) If x^2 + 12x + 20 < 0, how many integer values can x be?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Sol:

x^2+12x+20 = x^2+10x+2x+20 = x(x+10)+2(x+10) = (x+2)(x+10)

(x+2)(x+10) < 0

Roots of x= -10,-2

Ranges; x<-10, -10<x<-2, x>-2

For x>-2, let's say x=0; (x+2)(x+10) = 2*10=20. It is a positive

Thus,
x>-2 is +ve
-10<x<-2 is -ve; alternated from previous
x<-10 is +ve; alternated from previous

For "<" sign, we must consider only -ves.

-10<x<-2 befits the value of x that satisfy the inequality.
x can be -9,-8,-7,-6,-5,-4,-3

Count=7

Ans: "B"

how is x>-2?
i calculated x<-2, or x<-10.
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Retired Moderator Joined: 20 Dec 2010
Posts: 1463
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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Baten80 wrote:
fluke wrote:
banksy wrote:
27. (SC) If x^2 + 12x + 20 < 0, how many integer values can x be?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Sol:

x^2+12x+20 = x^2+10x+2x+20 = x(x+10)+2(x+10) = (x+2)(x+10)

(x+2)(x+10) < 0

Roots of x= -10,-2

Ranges; x<-10, -10<x<-2, x>-2

For x>-2, let's say x=0; (x+2)(x+10) = 2*10=20. It is a positive

Thus,
x>-2 is +ve
-10<x<-2 is -ve; alternated from previous
x<-10 is +ve; alternated from previous

For "<" sign, we must consider only -ves.

-10<x<-2 befits the value of x that satisfy the inequality.
x can be -9,-8,-7,-6,-5,-4,-3

Count=7

Ans: "B"

how is x>-2?
No, x is not greater than -2.

-10<x<-2 befits the value of x that satisfy the inequality.
x can be -9,-8,-7,-6,-5,-4,-3

Where I mentioned; "x>-2", I was just checking the side(+ve or -ve) of the inequality for the range where "x>-2"

i calculated x<-2, or x<-10.

I am afraid, this is not correct!!! It should be;
x<-2 AND x>-10

I have incorporated following method to calculate the range;

http://gmatclub.com/forum/inequalities-trick-91482.html

I humbly suggest you go through the same.
Manager  Joined: 24 Nov 2010
Posts: 152
Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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dreambeliever wrote:
banksy wrote:
27. (SC) If x^2 + 12x + 20 < 0, how many integer values can x be?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

x^2 + 12x + 20 < 0 --> (x+10)(x+2)<0 this will be true only when x<-2 and x> -10.

Ans = C

-10 to -2 exclusive on the number line leaves 7 integers not 8. have to quit committing this types mistakes!
Senior Manager  Joined: 01 Feb 2011
Posts: 491
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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solving the inequality

one solution x < -10 , x>-2 , unlimited integers can be chosen for these

second solution x >-10 , x<-2 , 7 possible integers ( -9,-8,-7...-3)

banksy wrote:
27. (SC) If x^2 + 12x + 20 < 0, how many integer values can x be?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Retired Moderator B
Joined: 16 Nov 2010
Posts: 1164
Location: United States (IN)
Concentration: Strategy, Technology
Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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The question asks, for how many integer values of x:

(x+10)(x+2) < 0

so -10 < x -2

only 7 integers between -10 and -2, so answer is B.
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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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After solving the equation, we get

(x+10)(x+2)<0

-10<x<-2

So, consider all the integer values within this range = -9, -8, -7, -6, -5, -4, -3

Which gives us 7 integer values, hence answer is B
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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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1
Top Contributor
banksy wrote:
If x² + 12x + 20 < 0, how many integer values can x be?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

GIVEN: x² + 12x + 20 < 0
Factor: (x + 2)(x + 10) < 0
In other words, (x + 2)(x + 10) is NEGATIVE

We can see that when x = -2 and x = -10, (x + 2)(x + 10) is ZERO
When we test some values, we can see the following:
If x < -10, then (x + 2)(x + 10) = (NEGATIVE)(NEGATIVE) = POSITIVE
If -10 < x < -2, then (x + 2)(x + 10) = (NEGATIVE)(POSITIVE) = NEGATIVE
If -2 < x, then (x + 2)(x + 10) = (POSITIVE)(POSITIVE) = POSITIVE

So, the only time (x + 2)(x + 10) is NEGATIVE is when -10 < x < -2
If x is an INTEGER, then x can equal -9, -8, -7, -6, -5, -4, and -3 (7 values)

Cheers,
Brent
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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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banksy wrote:
If x^2 + 12x + 20 < 0, how many integer values can x be?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Factoring, we have:

(x + 10)(x + 2) < 0

We see that when x is any integer from -9 to -3, inclusive, one of the factors is positive, and the other is negative, and thus the expression is less than zero. Since there are -3 - (-9) + 1 = 7 integers from -9 to -3, inclusive, there are 7 integer solutions.

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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?  [#permalink]

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banksy wrote:
If x^2 + 12x + 20 < 0, how many integer values can x be?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Asked: If x^2 + 12x + 20 < 0, how many integer values can x be?

(x+ 10)(x+2) < 0

-10 < x < -2

x = {-9,-8,-7,-6,-5,-4,-3}

IMO B
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Email: kinshook.chaturvedi@gmail.com Re: If x^2 + 12x + 20 < 0, how many integer values can x be?   [#permalink] 06 Jun 2020, 22:28

# If x^2 + 12x + 20 < 0, how many integer values can x be?  