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## If x is an integer, how many possible values

tagged by: Vincen

This topic has 3 expert replies and 0 member replies

### Top Member

Vincen Legendary Member
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07 Sep 2017
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#### If x is an integer, how many possible values

Tue Dec 05, 2017 9:31 am
If x is an integer, how many possible values of x exist for $$x^2+5\left|x\right|+6=0?$$
A. 4
B. 2
C. 3
D. 1
E. 0

The OA is E.

How can I conclude that E is the correct answer? I don't know how to prove it.

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### GMAT/MBA Expert

Matt@VeritasPrep GMAT Instructor
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Tue Dec 05, 2017 6:27 pm
I've got a quick way!

xÂ² + 5|x| + 6 = 0

(|x| + 2) * (|x| + 3) = 0

(|x| + 2) = 0 or (|x| + 3) = 0

|x| = -2 or |x| = -3

But no absolute values are negative, so there are no real solutions to this equation.

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### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Tue Dec 05, 2017 11:17 am
Vincen wrote:
If x is an integer, how many possible values of x exist for $$x^2+5\left|x\right|+6=0?$$
A. 4
B. 2
C. 3
D. 1
E. 0
Nice question!

IMPORTANT CONCEPT #1
xÂ² is greater than or equal to 0 for ALL values of x

IMPORTANT CONCEPT #2
|x| is greater than or equal to 0 for ALL values of x

So, xÂ² + 5|x| + 6 = 0 becomes... (some number greater than or equal to 0) + 5(some number greater than or equal to 0) + 6 = 0
Subtract 6 from both sides to get: (some number greater than or equal to 0) + 5(some number greater than or equal to 0) = -6
As you can see, there is now way that the left side of the equation can have a negative sum.
So, there are no solutions to the original equation.

Cheers,
Brent

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### GMAT/MBA Expert

ErikaPrepScholar Master | Next Rank: 500 Posts
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Tue Dec 05, 2017 9:56 am
We can solve this problem with logic and some number sense. As a refresher, we know that

positive * positive = positive
negative * negative = positive
positive * negative = negative
zero * anything = zero

Let's look at each term individually.

x^2: If x is an integer, this term will be either 0 (if x is 0) or positive (if x is negative or positive).
5|x|: 5 is positive, and the absolute value of x will be either 0 (if x is 0) or positive (if x is negative or positive). So this term will be either 0 or positive.
6: 6 is positive.

So the lowest possible number we can get from this equation is 6 (if x = 0). Any other number, negative or positive, will yield positive values for x^2 and 5|x|. This means there is no way for this equation to equal 0. So there are no possible values of x that make this equation true.

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