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.
Ron has been teaching various standardized tests for 20 years.
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