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M02-32

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Joined: 16 Jul 2017
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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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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 (1-x) = 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
msk0657 wrote:
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 (1-x) = 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.

x^2 could be 0 if x is 0. If we were told that x is not 0, then x^2 will be more than 0.
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31 Oct 2017, 20:14
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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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 2x-1>0 so we can have x^2=2x-1 then we can reach 1
therefore statement 1 is not sufficent
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04 Apr 2018, 07:51
umur wrote:
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 2x-1>0 so we can have x^2=2x-1 then we can reach 1
therefore statement 1 is not sufficent

My friend, it seems that you are absolutely wrong.

x = 0.4142 (I guess you wanted to say $$\sqrt{2}-1\approx 0.4142$$) does not satisfy $$x = \sqrt{2x - 1}$$. In this case 2x - 1 becomes negative. Even roots from negative numbers are not defined for the GMAT.

In addition to that, the way you are solving is wrong.
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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(2x-1) )^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 = (x-1)^2 = 0
b) x^2 + 2x + 1 = (x+1)^2 = 0

From a) you'd get |x-1| = 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(2x-1) )^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 = (x-1)^2 = 0
b) x^2 + 2x + 1 = (x+1)^2 = 0

From a) you'd get |x-1| = 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 non-negative x, |x| = x.
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27 Apr 2018, 03:46
Of course... Got it! Thanks a lot!!
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13 Dec 2018, 08:31
Bunuel In 1st statement,
what if I put value of x= 5
then root 2x-1= 9 and x=3 , what i am missing here?
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13 Dec 2018, 21:39
16SK16 wrote:
Bunuel In 1st statement,
what if I put value of x= 5
then root 2x-1= 9 and x=3 , what i am missing here?

x = 5 is not a solution of $$x = \sqrt{2x - 1}$$. The solution shows that the only root of the equation is x = 1. Not sure what you are trying to do there or even what you are trying to ask...
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04 Apr 2019, 03:44
I may be asking a very stupid question.
But When we square both side, should we not consider both cases: one with positive and the other with negative?
Case 1: x^2=2^x−1
Case 2: x^2=1-2^x

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04 Apr 2019, 04:10
Akash777 wrote:
I may be asking a very stupid question.
But When we square both side, should we not consider both cases: one with positive and the other with negative?
Case 1: x^2=2^x−1
Case 2: x^2=1-2^x

We have $$x = \sqrt{2x - 1}$$.

1. Even roots, such as the square root, is NOT defined for negative numbers on the GMAT. Thus 2x - 1 MUST be non-negative (0 or positive)
2. Even roots cannot give negative results. For example, $$\sqrt{9}=3$$ ONLY, not +3 and -3. Thus, x, which equals to $$\sqrt{2x - 1}$$, cannot be negative.

So, when you square and solve for x, you should check that 2x - 1 and x are positive or 0. That's it.
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