if a is -1/2 then b cannot be +1/2, as b has to be a number between modulus a and 1. modulus a will be +1/2 if a is -1/2.

Aint it wrong cosidering bigger value of a all the time?

sir, if we divide by ab on both side , we get, i am referring to third simplified form , QA: b*a^3 and QB: ab^3,
QA QB
a^2 b^2
then we have to take square root on both side ?

Can someone please help me understand this,

Simplifying everything, I got Quantity A : a^2
Quantity B : b^2

I picked a = -0.5 and b = 0.9

In this case B is greater as 0.81 > 0.25

Quantity A: \(a^3b\), where \(a^3>0\)
Quantity B: \(ab^3\)

cancelling out \(ab\) results in
Quantity A: \(a^2\)
Quantity B: \(b^2\)

Since \(a>-1\), and \(0<b<1\), answer is A

Use the placeholder method here
Simplify it to obtain a^3b ? ab^3
Since a<0, the placeholder sign will flip. Now use a flip operator on the placeholder sign, say it is now ?' after the flip
a^2 ?' b^2, since |a|<b, a^2 < b^2
Hence, ?' is <, so ? must be >, hence A>B and that is the answer.

In the left equation, it has a square root of A, which is negative.
I think it will require a use of the imaginary numbers...

Great point!
I didn't consider that while composing my solution.
Given that, I don't think this could ever be an official GRE question.

The question if the question tests the right concepts for the exam or less. IF the answer is not...just skip the question sir

Hello,

I am new in this forum. I believe this question has a fundamental issue: By definition, negative numbers don't have square roots in the real numbers' domain. So, I am tempted to choose D as answer since [square_root][a] is undefined (a is a negative number).

Cheers
Ariel

Yes sir

brent pointed out also above

I think the problem is invalid, because any square root function is defined only when the input is a non-negative number. In the left-hand side, there is a square root of a, but a is less than zero, so it will be invalid.

This concept does not apply to the square roots of negative numbers, which is the case here.

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