you can't just take square roots of even powers, unless you have previous assurance that the variables represent positive numbers.
otherwise you must use absolute values: √(n^2) = |n|.

so, the first of these statements gives |a| = |b|, which implies a = ±b.
the second gives |a| = |b^2|, or |a| = b^2, implying a = ±(b^2).

sumgb wrote:my take on this problem is ...

stmnt 1, a=b=1, true; a=b=-1 false so insuff.

hmm? how do you think it's false for a = b = -1?
if you plug a = b = -1 into the statement, then it's definitely true. (in fact, since you doing exactly the same thing to both numbers -- in this case, squaring them -- it should be clear that the statement will always be true if a and b are the same number.)
it's also true for a = 1, b = -1, and vice versa.

stmnt 2, since a^2 = b^4 this implies a = b^2 (taking square root) which means a is +ve since b^2 cant be negative.

nope.
see above.
also plug in a = b = 1;
a = b = -1;
a = 1, b = -1;
and a = -1, b = 1, and you'll quickly notice that all of them work.

the answer should be (e); this is an error in the book. it will be fixed in the next edition.

gmat25 wrote:Now, as we simplify the eq'n in most of the DS question, i did the same here. Multiply the above eq'n with "a" on both sides, so u get,

1 > a^2 / (b^4 + 3)

this step is wrong; you can't multiply an inequality by a variable of unknown sign.

if you do that, then you get ">" if a turns out to be positive, but you get "<" if a turns out to be negative.
if you want to address each of these two different inequalities according to the possibilities for a, then that approach will also work -- but it's more laborious than leaving the inequality in its current form.

to prove that your algebra is incorrect, plug a = 1, b = 1 and a = -1, b = -1 into the ORIGINAL question, and you'll see that you get one "yes" and one "no".

sumgb wrote:

hmm? how do you think it's false for a = b = -1?

Hi Ron,
If I use a=b= -1 in the expression 1/a > a/(b^4+3); it becomes -1 > -1/4 which is a NO (not false, my mistake). whereas a=b=1 you get an answer YES.

ah, i thought you were saying that statement (1) itself was false if you input these numbers. ok, got it.

TOPGMAT wrote:If a does not equal to zero, is 1/a > a/(b^4 +3)?

(1) a^2=b^2

(2) a^2=b^4


Can we do it this way?
if a>0 => 1 > a^2/(b^4+3) and assume a=1,b=1 => True.
if a<0 => 1 < a^2/(b^4+3) and assume a=-1,b=1 => 1 < 1/4 => False.

yep -- that works.