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x^3 < x^2 Describe the possible values of x

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by gmattesttaker2 » Sun Jun 22, 2014 8:56 pm
Hello,

For the following:

x^3 < x^2. Describe the possible values of x.

Is this correct?

x^3 < x^2
=> x^3 - x^2 < 0
=> x^2(x - 1) < 0

Since x^2 > 0 => x - 1 < 0 => x < 1 - I

x^2 > 0 = sq. root (x^2) > sq. root (0)
=> +/- x > 0
=> x > 0 or -x > 0
=> x > 0 or x < 0 - II

However, from I and II, I was not sure what the final answer will be.

Can you please help? The answer given is Any non-zero number less than 1

Thanks a lot,
Sri
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by [email protected] » Sun Jun 22, 2014 9:16 pm
Hi Sri,

Rather than go about this algebraically, you can TEST VALUES to figure out what's "possible" and what's "not"

X^3 < X^2

Let's start with the obvious: POSITIVE INTEGERS. Will ANY positive integers "fit" this inequality?

X = 1 gives us 1 < 1; this does NOT fit
X = 2 gives us 8 < 4; this does NOT fit

As X becomes a larger positive integer, we run into the same problem: X^3 is NOT < X^2. We've eliminated a set of values. X CANNOT be greater than or equal to 1.

Next, let's try NEGATIVE INTEGERS. There's a great Number Property here...

(Negative)^3 = Negative
(Negative)^2 = Positive

From this, we can deduce that ANY NEGATIVE (integer or non-integer) will "FIT" this inequality.

We're not done though. We still have to deal with 0 and POSITIVE FRACTIONS.

If X = 0, then we have 0 < 0; this does NOT fit.

If X = 1/2, then we have 1/8 < 1/4; this DOES fit. We can also make a deduction from this...

Cubing a positive fraction that is less than 1 IS less than squaring that same positive fraction.

From these TESTS, we've deduced the following:
X CANNOT be a positive integer nor can it be greater than 1
X CANNOT be 0

X CAN be ANY negative
X CAN be ANY positive fraction less than 1

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by GMATGuruNY » Mon Jun 23, 2014 2:26 am
If x³ < x², describe all possible values of x.
The inequality above implies that x≠0.
Thus, whether x is positive or negative, x²>0, implying that we can safely divide each side by x²:
x³/x² < x²/x²
x < 1.
Thus, x can be equal to any nonzero value less than 1.
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by GMATinsight » Mon Jun 23, 2014 8:49 pm
@gmattesttaker2

You didn't have to do any calculation as mentioned below

"Since x^2 > 0 => x - 1 < 0 => x < 1 - I

x^2 > 0 = sq. root (x^2) > sq. root (0)
=> +/- x > 0
=> x > 0 or -x > 0
=> x > 0 or x < 0 - II "

This calculation was a futile exercise because this part has to be positive for any value of x and therefore doesn't lead us to any conclusion except that x should be non zero. Probaly because you did this calculation therefore you got confused.

The first calculation already gave you that x<1 and x has to be non xero for defining the given function therefore the answer you got in half of your exercise already which is "X can be any non zero number less than 1"

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