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OG #154

Expert replies
by Stockmoose16 » Sat Oct 04, 2008 5:35 pm
Hello,

This question has been talked about before on this forum, but I don't understand why my method isn't working:

Is x negative?

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

(2) x^2-1<0



1) Distribute the exponent:

X^3-X^5<0

Move the terms to opposite sides of the equation

X^3<X^5

Test numbers:

If X >1 (i.e. X=2, X^3=8, X^5=32) -- stays true to given equation

If 0>X>-1 (i.e. X= -1/2, X^3= -1/8, X^5= -1/32) -- does not work with given equation, since X^3>X^5

Thus... this statement is SUFFICIENT, because only positive numbers work...

The O.G. says that any positive number greater than 1 will work, as will any negative number between 0 and -1. But I just proved that any number between 0 and -1 does not stay true to the given equation. What's wrong with the way I approached this question?
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Source: — Data Sufficiency |

Re: OG #154

by LSB » Sat Oct 04, 2008 5:55 pm
Stockmoose16 wrote:Hello,

This question has been talked about before on this forum, but I don't understand why my method isn't working:

Is x negative?

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

(2) x^2-1<0



1) Distribute the exponent:

X^3-X^5<0

Move the terms to opposite sides of the equation

X^3<X^5

Test numbers:

If X >1 (i.e. X=2, X^3=8, X^5=32) -- stays true to given equation

If 0>X>-1 (i.e. X= -1/2, X^3= -1/8, X^5= -1/32) -- does not work with given equation, since X^3>X^5
X^3 < X^5
-1/8 < -1/32

These equations are consistent. Hence X can be negative or positive (as you have shown)
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