If x is an integer, how many possible values

[Vincen]

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. <i class="em em-cry"></i>

[ErikaPrepScholar]

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.

[Brent@GMATPrepNow]

If x is an integer, how many possible values of x exist for x² + 5|x| + 6 = 0?
A. 4
B. 2
C. 3
D. 1
E. 0

Nice question!

IMPORTANT CONCEPT #1
x² 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, x² + 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.

Answer: E

Cheers,
Brent

[Matt@VeritasPrep]

I've got a quick way!

x² + 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.