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If x^2 + 12x + 20 < 0, how many integer values can x be?
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17 Mar 2011, 13:44
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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
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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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17 Mar 2011, 14:32
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



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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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17 Mar 2011, 18:30
the same solution and also found 8<a<2 and C is the answer



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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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17 Mar 2011, 22:34
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"



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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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18 Mar 2011, 07:34
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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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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18 Mar 2011, 08:21
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. Please read my post closely; 10<x<2 befits the value of x that satisfy the inequality. x can be 9,8,7,6,5,4,3Where 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/inequalitiestrick91482.htmlI humbly suggest you go through the same.



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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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18 Mar 2011, 10:32
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!



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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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18 Mar 2011, 16:54
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) Hence the answer is B. 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



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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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18 Mar 2011, 21:57
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?
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19 Jun 2017, 13:37
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?
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22 Nov 2019, 16:21
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 NEGATIVEWe 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) Answer: B Cheers, Brent
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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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27 Nov 2019, 11:00
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. Answer: B
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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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06 Jun 2020, 22:28
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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Re: If x^2 + 12x + 20 < 0, how many integer values can x be?
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