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Re M0232
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30 May 2017, 05:59
I think this is a poorquality question and I don't agree with the explanation. The question asks the value of x^2. Why do we calculate x and then x^2. If X^2 is the value required, both statements independently be sufficient.



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30 May 2017, 08:06



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17 Jul 2017, 02:21
I don't agree with the explanation. the answer should be D .If we divide both sides of the 2nd equation with x^2 , we are left with 1=x , thus x^=1



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17 Jul 2017, 02:29
Arjun17 wrote: I don't agree with the explanation. the answer should be D .If we divide both sides of the 2nd equation with x^2 , we are left with 1=x , thus x^=1 It's always better to read the whole discussion before posting a question. You are making a rookie mistake, which is addressed in a post just above yours. You cannot divide x^2 = x^3 by x^2 because x^2 could be 0 and we cannot divide by 0. By doing so, you are loosing a root, namely x = 0 (notice that both x = 0 and x = 1 satisfy x^2 = x^3, and this is clearly shown in the original solution). Takeaway: never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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Re: M0232
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03 Oct 2017, 11:27
Bunuel,
Can you please throw some light on my query.
For statement 2: I made till this point \(x^2\) (1−x)=0  > eq 1
When I segregated \(x^2\) = 0 and (1x) = 0. Since \(x^2\) is always +ve and can't be 0 , so I eliminated this part. Finally, got x = 1.
I have seen such type of questions,even though when \(x^2\) is 0 some times \(x^2\) taken +ve and in few solutions it is not( case in our question). Under what cases we need to consider +ve or 0 when we have such eq 1 type.



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03 Oct 2017, 11:30



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Quick question, from 1, because we are not given x=/= 0 in the stem, we know we can square the equation because 0 =/= (1)^(1/2), b/c we cannot take root of 1, is that correct? Is there a general way to check whether we are able to perform operations on variables in equations and inequalities? Thanks in advance.



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31 Oct 2017, 21:10
alekkx wrote: Quick question, from 1, because we are not given x=/= 0 in the stem, we know we can square the equation because 0 =/= (1)^(1/2), b/c we cannot take root of 1, is that correct? Is there a general way to check whether we are able to perform operations on variables in equations and inequalities? Thanks in advance. Check : Manipulating Inequalities (adding, subtracting, squaring etc.). For more check Ultimate GMAT Quantitative Megathread
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Re: M0232
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04 Apr 2018, 07:42
the quesiton is absolutely wrong statement 1 does not satisfy the direct solution there can be two ways to solve the equation if 2x 1<0 then x^2=2x+1 so we can have o positive root for x which is 0.4142 if u substitute in eq u will see and the other one is if 2x1>0 so we can have x^2=2x1 then we can reach 1 therefore statement 1 is not sufficent



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04 Apr 2018, 07:51



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27 Apr 2018, 02:46
Hi Bunnel,
Quick question:
I learned from gmatclub exercises that, whenever the sign of a variable is unknown, we should apply the module when we square a square root.
so in this exercise, from (1) you get:
x^2 = ( sqroot(2x1) )^2
shouldn't it be:
x^2 = 2x  1, which would yield x^2 = 2x  1 or x^2 = 2x + 1 and thus we would get 2 quadratic expressions:
a) x^2 2x + 1 = (x1)^2 = 0 b) x^2 + 2x + 1 = (x+1)^2 = 0
From a) you'd get x1 = 0 and hence x = 1 From b) you'd get x+1 = 0 and hence x= 1
but since we know from the stem that x = sqroot of something, hence x has to be > 0 and thus b) is discarded and the statement is sufficient.
My question is, in your solution you have just squared the root and took the positive value straight ahead and your process was much more simple and quick. Why?



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27 Apr 2018, 03:44
Miracles86 wrote: Hi Bunnel,
Quick question:
I learned from gmatclub exercises that, whenever the sign of a variable is unknown, we should apply the module when we square a square root.
so in this exercise, from (1) you get:
x^2 = ( sqroot(2x1) )^2
shouldn't it be:
x^2 = 2x  1, which would yield x^2 = 2x  1 or x^2 = 2x + 1 and thus we would get 2 quadratic expressions:
a) x^2 2x + 1 = (x1)^2 = 0 b) x^2 + 2x + 1 = (x+1)^2 = 0
From a) you'd get x1 = 0 and hence x = 1 From b) you'd get x+1 = 0 and hence x= 1
but since we know from the stem that x = sqroot of something, hence x has to be > 0 and thus b) is discarded and the statement is sufficient.
My question is, in your solution you have just squared the root and took the positive value straight ahead and your process was much more simple and quick. Why? The point is that \(\sqrt{x^2}=x\) but \((\sqrt{x})^2=x\). Here x is under the square root so it cannot be negative, and it does not makes sense to write x instead of x, because for nonnegative x, x = x.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M0232
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27 Apr 2018, 03:46
Of course... Got it! Thanks a lot!!







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