I am just curious, is that math mathematically valid? I might be terribly wrong. Even that, I wanna share my thought. As we know a must be negative. So in A, how could we validate the square root of something negative?


Thanks,

a is negative because is between -1 and zero.

Must be negative.

However, when it is inside the absolute value, it must be positive. After zero.

regards

You're absolutely correct. It's a bad question.
I didn't look closely inside the bracketed part, since it all gets squared in the end.
If a < 0, then √a is undefined, which renders the question invalid.
The GRE would never have a question like this.

Cheers,
Brent

Carcass GreenlightTestPrep

I got the answer as B in this way. I learnt comparison in this way from Kaplan. Is this valid?

given quantity A :

(a^2 *√b)^2 = (a^4 * b)\ a = a^3 b
___________
(√a)^2


given quantity b :

a * b^5
_______ = ( a * b^5) / b^2 = a * b^3
(√b)^4

comparing qty A and qty B

a^3 * b = a * b^3

simplifying by dividing both sides by ab we get

Qty A = a^2

Qty B = b ^2

we are given a range to pick values for a and b

so when I take a = -0.6 and Mod a = 0.6 which must be less than b. So I pick b = 0.8 and b must be less than 1.

now squaring a and b

I get Qty A = 0.36 and Qty B = 0.64

Now B is the answer isn't it?

Pls correct me if am wrong somewhere.

Hi

the answer is A and I suggest you to read very careful the GreenlightTestPrep explanation.

it is neat and marvelous.

Also read our book for better insights https://gre.myprepclub.com/forum/gre-math- ... -2609.html

Feel free to ask for further assistance

PS: I saw the one you posted method pretty cumbersome

This isn't a good question, and it won't help you prepare for the GRE.
Since \(a\) is negative, \(\sqrt{a}\) is not defined. In other words, it has no real value.

This means evaluating \(\left[ \begin{array}{cc|r} \frac{a^2 \sqrt{b}}{ \sqrt{a}} \end{array} \right]^2\) is similar to evaluating \(\left[ \begin{array}{cc|r} \frac{a^2 \sqrt{b}}{apple} \end{array} \right]^2\)

Hi, i got the same as yours, but what i figure out in last is we cannot divide the quantity if it is negative. product of ab is clearly negative thats why we misscalculate the question.

Post locked

The discussion is located here https://gre.myprepclub.com/forum/1-a-0-a-b ... tml#p23077

Regards