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find the number of integers satisfying the inequality (x^2 +

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find the number of integers satisfying the inequality (x^2 + [#permalink] New post 05 Jun 2006, 09:34
find the number of integers satisfying the inequality (x^2 + x - 6) (x^2 - x - 20) < 0

a. 4 b.2

c.0 d. infinitely many
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Re: inequalities and modulus [#permalink] New post 05 Jun 2006, 10:18
yasmeen wrote:
find the number of integers satisfying the inequality (x^2 + x - 6) (x^2 - x - 20) < 0

a. 4 b.2

c.0 d. infinitely many


b - 2

only 3, 4 hold true
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 [#permalink] New post 05 Jun 2006, 11:27
I'll go with A - 4 integers


Break this equation out: (x^2 + x - 6) (x^2 - x - 20) < 0

(x+3)(x-2)(x-5)(x-4)<0

either one or three () must be a negative integer in order for the equation to be less than 0.

If x>5, then the equation is positive
If x=4, 5, 2 then the equation is 0
If x=3, the equation is positive (two positives*two negatives)

X can equal 1, 0, -1, -2

If x<-2, then the equation is positive or 0 (when x=-3)

Therefore that are 4 possibilities for x.
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Re: inequalities and modulus [#permalink] New post 05 Jun 2006, 11:40
yasmeen wrote:
find the number of integers satisfying the inequality (x^2 + x - 6) (x^2 - x - 20) < 0

a. 4
b.2
c.0
d. infinitely many


A. 4 (the numbers are 1, 2, 3 and 4). if we donot consider -ves.
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 [#permalink] New post 05 Jun 2006, 12:35
B.

(x-2)(x+3)(x+4)(x-5)<0

only if x=3 or 4.
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 [#permalink] New post 05 Jun 2006, 13:06
tl372 wrote:
I'll go with A - 4 integers


Break this equation out: (x^2 + x - 6) (x^2 - x - 20) < 0

(x+3)(x-2)(x-5)(x-4)<0

either one or three () must be a negative integer in order for the equation to be less than 0.

If x>5, then the equation is positive
If x=4, 5, 2 then the equation is 0
If x=3, the equation is positive (two positives*two negatives)

X can equal 1, 0, -1, -2

If x<-2, then the equation is positive or 0 (when x=-3)

Therefore that are 4 possibilities for x.


How did you conclude this??

If x=0, then expression (x^2 + x - 6) (x^2 - x - 20) = (-6).(-20) >0
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 [#permalink] New post 05 Jun 2006, 13:24
(B) as well

x = 3 and 4.
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Re: inequalities and modulus [#permalink] New post 05 Jun 2006, 13:26
Professor wrote:
yasmeen wrote:
find the number of integers satisfying the inequality (x^2 + x - 6) (x^2 - x - 20) < 0

a. 4
b.2
c.0
d. infinitely many


A. 4 (the numbers are 1, 2, 3 and 4). if we donot consider -ves.


prof, x=1 does not make the expression < 0

or am i missing something, can you please re-evaluate?
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 [#permalink] New post 05 Jun 2006, 13:32
gmatmba wrote:
tl372 wrote:
I'll go with A - 4 integers


Break this equation out: (x^2 + x - 6) (x^2 - x - 20) < 0

(x+3)(x-2)(x-5)(x-4)<0

either one or three () must be a negative integer in order for the equation to be less than 0.

If x>5, then the equation is positive
If x=4, 5, 2 then the equation is 0
If x=3, the equation is positive (two positives*two negatives)

X can equal 1, 0, -1, -2

If x<-2, then the equation is positive or 0 (when x=-3)

Therefore that are 4 possibilities for x.


How did you conclude this??

If x=0, then expression (x^2 + x - 6) (x^2 - x - 20) = (-6).(-20) >0



My mistake...was careless when I wrote down the equation. Should be:

(x+3)(x-2)(x-5)(x+4)<0

Therefore only two solutions 3, and 4.
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Re: inequalities and modulus [#permalink] New post 05 Jun 2006, 16:33
gmatmba wrote:
Professor wrote:
yasmeen wrote:
find the number of integers satisfying the inequality (x^2 + x - 6) (x^2 - x - 20) < 0
a. 4
b.2
c.0
d. infinitely many

A. 4 (the numbers are 1, 2, 3 and 4). if we donot consider -ves.

prof, x=1 does not make the expression < 0
or am i missing something, can you please re-evaluate?


nope. i was missing. it is B.

the only satisfying integers are 3 and 4 not 1, 2, 0, -1, -2, -3, -4, -5 and so on.
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Re: inequalities and modulus [#permalink] New post 05 Jun 2006, 22:09
Professor wrote:
gmatmba wrote:
Professor wrote:
yasmeen wrote:
find the number of integers satisfying the inequality (x^2 + x - 6) (x^2 - x - 20) < 0
a. 4
b.2
c.0
d. infinitely many

A. 4 (the numbers are 1, 2, 3 and 4). if we donot consider -ves.

prof, x=1 does not make the expression < 0
or am i missing something, can you please re-evaluate?


nope. i was missing. it is B.

the only satisfying integers are 3 and 4 not 1, 2, 0, -1, -2, -3, -4, -5 and so on.


Have detailed the way I solved the problem. Might be useful.
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Re: inequalities and modulus [#permalink] New post 06 Jun 2006, 04:30
shobhitb wrote:
Professor wrote:
gmatmba wrote:
Professor wrote:
yasmeen wrote:
find the number of integers satisfying the inequality (x^2 + x - 6) (x^2 - x - 20) < 0
a. 4
b.2
c.0
d. infinitely many

A. 4 (the numbers are 1, 2, 3 and 4). if we donot consider -ves.

prof, x=1 does not make the expression < 0
or am i missing something, can you please re-evaluate?


nope. i was missing. it is B.

the only satisfying integers are 3 and 4 not 1, 2, 0, -1, -2, -3, -4, -5 and so on.


Have detailed the way I solved the problem. Might be useful.


:good Good work shobhit. Interval method is the best for solving both Modulus and polynomial inequality questions.
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Re: inequalities and modulus   [#permalink] 06 Jun 2006, 04:30
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