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.

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.

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 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

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.