Michael,

I think you are making it more complicated that it actually is. lets see:

for statement 1: if x<0

that means x*abs(x) is negative. Where as 2^x is a positive value because 2^(-x) = 1/2^x which is positive for any negative x. hence x*abs(x) < 2^x --- sufficient

Statement 2: x=-10 --- this is same as statement 1 (x is negative) but instead giving us definite value. Therefore sufficient.

Answer D.

Hope this helps

^ Right, I understand that.

However, my first instinct for these types of problems is to break it down to 2 equations to make it. For example, is |x| < 1, my first step is always: x<1 and x>-1. Making it: -1 < x < 1

My question is, why won't this approach work for this particular problem?

In given scenario it sufice to note whether x<0 as x|x| < 0 for x<0 while 2^x will always be positive.

michaelfaulkner wrote:I can't figure out, for an absolute value problem, if and when I should separate the equation into two or more equations when testing cases.

For example, look at this problem:

is x|x| < 2^x?
(1) x < 0
(2) x=-10

So for item (1), if I test negative cases directly into the equation, I get yes on every instance. However, my first instinct for these types of problems is to break the absolute value equation into two equations:

x|x| < 2^x equals...

x^2 < 2^x AND -x(x) > 2^x

Your approach illustrates one of the dangers of using algebra when solving a problem with absolute values. The inequality -x(x) > 2^x has no solutions. The left side will always be negative (unless x=0, in which case the left side will equal 0); the right side will always be positive. So the left side can never be greater than the right side.

Here's a much safer approach:

Statement 1: x<0
Plugging in a negative value into x|x| < 2^x, we get:
negative * positive < positive.
negative < positive.
Sufficient.

Statement 2: x= -10
Since in statement 1 we discovered that x<0 is sufficient, knowing that x= -10 also is sufficient.

The correct answer is D.

^Thank you for the great response.

So is it safe to assume that it is usually always a bad idea to use algebra to solve for data sufficiency absolute value questions?

michaelfaulkner wrote:^Thank you for the great response.

So is it safe to assume that it is usually always a bad idea to use algebra to solve for data sufficiency absolute value questions?

I would tread carefully. The problem above involves a quadratic, making the use of algebra especially tricky.

When there's no quadratic, algebra can be helpful:

Is |2x+5| > |3x-2|?

Let's use algebra to find the points of intersection:
2x+5 = 3x-2
x = 7

2x+5 = -(3x-2)
2x+5 = 2-3x
5x = -3
x = -3/5

The points of intersection (x=7 and x= -3/5) are where the situation will change:

When x>7:
Plug in x=8:
|2x+5| = |2*8 + 5| = 21.
|3x-2| = |3*8 - 2| = 22.

This shows us that when x>7, |2x+5| < |3x-2|

When -3/5 < x < 7:
Plug in x=0:
|2x+5| = |2*0 + 5| = 5.
|3x-2| = |3*0 - 2| = 2.

This shows us that when -3/5 < x < 7, |2x+5| > |3x-2|.

When x<-3/5:
Plug in x= -1
|2x+5| = |2*-1 + 5| = 3.
|3x-2| = |3*-1 - 2| = 5.

This shows us that when x < -3/5, |2x+5| < |3x-2|.

Since |2x+5| > |3x-2| only when -3/5 < x < 7, the question above can be rephrased:

Is -3/5 < x < 7?

Using algebra helps us to rephrase the question, and rephrasing the question makes it much easier to evaluate the two statements. So algebra can certainly be helpful. Just be careful!