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How many of the integers that satisfy the inequality (x+2)(x

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Intern
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Status: GMAT in August 2018
Joined: 05 Mar 2018
Posts: 46
Location: India
Concentration: Leadership, Strategy
WE: Law (Consulting)
Re: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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New post 19 Apr 2018, 02:46
My simple way to solve: Experts can suggest if its flawed.

(x+2)(x+3)/(x-2) ≥ 0

Try make a square in the denominator by multiplying both numerator and denominator by (x-2).

Then equation would like like this: (x+2)(x-2)(x+3)/(x-2)^2 ≥ 0

The denominator shall be +ive for all values of x, other than 2, where the expression will become undefined. Remember to exclude x=2 in all solutions.

Coming to numerator, (x+2)(x-2) makes (x^2-2^2) => (x^2-4). This will be = 0 or +ive for for x>= 2 and x<= -2. So list of possible solutions of x = 2,3,4..... and -2, -3, -4.... Remember to exclude x=2 here.

Coming to the other remaining term in the numerator (x+3) => for this to be 0 or +ive, x can be any value of x >= -3. i.e. -3, -2, -1.....

Now observe what values are common across all solution = -3, -2 and 3, 4, for the question only wants values less than 5.

Cheers!
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Re: How many of the integers that satisfy the inequality (x+2)(x  [#permalink]

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New post 21 Aug 2018, 21:04
cyberjadugar wrote:
Hi,

General method:

\({(x+2)(x+3)}/(x-2) \geq 0\)

if we plot it on number line, we have,
\(-3 \leq x \leq -2\)
& \(x > 2\), since \(x-2 \neq 0\) (no equality).

Also, it is given\(x < 5\)
Thus integral solutions would be x = -3, -2, 3, 4

Answer is (D)


Hi CyberJadugar,

Could you please explain how you got 3 & 4 as the integral solutions.

I understand how you get -3 & -2 as the answers but 3 & 4 does not make sense to me. As the question is asking for the final value not to be greater than 5.

Regards,
GMAT Club Bot
Re: How many of the integers that satisfy the inequality (x+2)(x &nbs [#permalink] 21 Aug 2018, 21:04

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