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

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Intern
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find the number of integers satisfying the inequality (x^2 + [#permalink]  05 Jun 2006, 09:34
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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
Manager
Joined: 09 May 2006
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Re: inequalities and modulus [#permalink]  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
Manager
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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.
VP
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Re: inequalities and modulus [#permalink]  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.
Manager
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B.

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

only if x=3 or 4.
Director
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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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SVP
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(B) as well

x = 3 and 4.
Director
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Location: France
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Kudos [?]: 10 [0], given: 0

Re: inequalities and modulus [#permalink]  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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Manager
Joined: 10 May 2006
Posts: 186
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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.
VP
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Re: inequalities and modulus [#permalink]  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.
Manager
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Re: inequalities and modulus [#permalink]  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.
Attachments

solution.doc [22.5 KiB]

VP
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Re: inequalities and modulus [#permalink]  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 work shobhit. Interval method is the best for solving both Modulus and polynomial inequality questions.
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- Bernard Edmonds

Re: inequalities and modulus   [#permalink] 06 Jun 2006, 04:30
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