Data Sufficiency

jainrahul1985 wrote:If x is negative, is x < -3? (1) x^2 > 9 (2) x^3 < -9

OA A

(1) x² can be greater than 9 if x > 3 or x < -3 because 3² = 9 and (-3)² = 9. Since x is negative, so the only possible value of x = -3; SUFFICIENT.

(2) x^3 < -9
If x = -2.1, then x^3 = -9.2 < -9; here x > -3
If x = -4, then x^3 = -64 < -9; here x < -3
No definite answer; NOT sufficient.

The correct answer is A.

i received a PM regarding this thread.

anurag --

Anurag@Gurome wrote:(1) x² can be greater than 9 if x > 3 or x < -3 because 3² = 9 and (-3)² = 9. Since x is negative, so the only possible value of x = -3; SUFFICIENT.

this is a correct solution of x^2 > 9 --> x>3 or x<-3.
however, the second part ("the only possible value of x = -3") is incorrect; the possible values of x are all values for which x < -3, i.e., all points to the left of -3 on the number line.

(2) x^3 < -9

with ODD powers and roots -- even in inequalities -- you can just do the powers and roots; you don't have to worry about positives/negatives or double cases (as you would with quadratics and other even powers).
therefore,
x < -(cube root of 9)
the cube root of 9 is just barely more than 2 (because the cube root of 8 is exactly 2); let's say about 2.1.
so, this statement just means x < -2.1.
that's not good enough to establish whether x < -3; see the specific cases listed by other posters for concrete proof.

lunarpower wrote:
this is a correct solution of x^2 < 9 --> x>3 or x<-3.

You surely meant to write "x^2 > 9" here.

Ian Stewart wrote:You surely meant to write "x^2 > 9" here.

sure did ... dang alligators, always turning around on me.

i edited that, so you can kill this post now if you want. thanks.