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DebbieChats
(1)+(2) Both a = -4 and a = 3 satisfy |a| > 2, so we cannot find the single numerical value of a. Not sufficient.

1+2
a< -2 and a=-4 ...so a is -4? (C) wrong logic?

We don't have that a < -2. We have that |a| > 2, which means that a < -2 OR a > 2. Again, BOTH value from (1), a = -4 and a = 3, satisfy |a| > 2.
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What is the value of a?

(1) a^2 + a – 12 = 0

(2) |a| > 2



Why is it not C?

(1) Factoring gives (a+4) (a-3) = 0 so a=-4 or a=3

NS

(2) |a| >2 --> a>2 or a<-2 NS

(1) and (2) a=-4 or a =3 and a <-2 or a>2 NS E
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What is the value of a?

(1) a^2 + a – 12 = 0

(2) |a| > 2



Why is it not C?


#1
a can be -4 or +3
not sufficient
#2
lal>2 not sufficient
from 1& 2
a can be any value -4 or 3
not sufficient
IMOE
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