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>
If x is an integer, how many possible values
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- ErikaPrepScholar
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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.
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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- Brent@GMATPrepNow
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Nice question!Vincen wrote: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
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
Last edited by Brent@GMATPrepNow on Thu May 09, 2019 5:01 pm, edited 1 time in total.
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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.
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.