voodoochild wrote:
VeritasPrepKarishma wrote:
Here your question is:
Is \(\sqrt{x} - \sqrt{y}> \sqrt{(x+y)}\)?
Mind you, it is not given to you that the inequality holds. You need to find out whether it does. The LHS can be negative here (in case root x is less than root y) on its own. In that case, the inequality certainly doesn't hold.
Karishma,
Thanks for your reply. Your replies are always helpful. However, I have a different opinion for two reasons:
#1- We are assuming that the given inequality is TRUE, and then evaluating its correctness. Hence, if x and y are positive, sqrt(x)-sqrt(y) will be positive. If it's -ve then the inequality cannot hold good. That's what happened in the end.
#2 - One of the best strategies on MUST BE TRUE questions is to solve the equation as much as possible. There are questions on which plugging-in numbers won't work at all.
Thoughts?
Having interacted with you over the past few months, I think that you genuinely care about having strong concepts. Hence, I will try to explain in detail the points you are missing (in my opinion):
#2 - One of the best strategies on MUST BE TRUE questions is to solve the equation as much as possible. There are questions on which plugging-in numbers won't work at all.
Absolutely! I dislike number plugging. For most questions, I follow the strategy of thinking logically; I try to apply conceptual understanding to questions. That said, in some questions, number plugging is what works. For 'MUST BE TRUE' questions, number plugging is useless if the relation actually must be true. You cannot plug in every number and check to establish that the relation is indeed true. So once you get that the relation holds for numbers from different ranges, you need to start thinking logically why it must hold for all numbers. On the other hand, if you can find one number for which the relation doesn't hold true, you are done! You already know that the relation doesn't hold for all numbers. Hence, in some cases, the most successful strategy is when you can see that the relation does not hold for an easy number e.g. 0 or 1 etc. Hence, number plugging has its uses, limited they may be.
In the context of this question, what I see is that you are removing the roots and then plugging in numbers (in statement II). Removing the roots didn't actually give you the range, right? Statement III worked out well but it was a fluke. I will explain this more in your next point.
#1- We are assuming that the given inequality is TRUE, and then evaluating its correctness. Hence, if x and y are positive, sqrt(x)-sqrt(y) will be positive. If it's -ve then the inequality cannot hold good. That's what happened in the end.
Fine. Let's assume that the inequality holds (statement III). We get that LHS must be positive. We do some algebra and we get that -2sqrt(xy)>0 which we know is not possible. This proves to us that the inequality definitely does not hold i.e. LHS is not greater than RHS for any value of x and y (x and y are both positive). The reason why this worked out is that the inequality does not hold for any value of x and y i.e. it is definitely false. It is possible that you get an inequality which holds for some values and does not for others. In that case, you would again need to plug in numbers and check (like you did in statement 2). Our question was whether the inequality holds for all values. In statement 3, by assuming that it does, we were able to prove that it doesn't hold for any value. The strategy fails in case it holds for some values and does not for others (like it did for statement 2).
Now, what irks me about assuming that the inequality holds and squaring? Let me show you using a simple example.
Say, my question is: Is A > B?
(A and B are expressions using x and y. Say, the values A can take are A>2 or A < -2 and the values B can take are 0 < B < 2. But of course, this is not given to you and the expressions are complex so you cant really see it either)
You assume that A > B and square it.
A^2 > B^2
Not you plug in values for x and y in the inequality and you see that for every value you put in, A^2 is greater than B^2 (it will be because of the range of values A and B can take). What does it imply? Does it mean that A > B for every value of x and y? But it is not true. A is not greater than B for every value of x and y. Hence, the entire exercise could be misleading.
Bottom line: Try to avoid assuming that the relation questioned actually holds true. (It might work splendidly in rare situations but more often than not, it will confuse you to no end)
_________________
Karishma
Veritas Prep GMAT Instructor
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