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If x is an integer, what is the value of x? [#permalink]
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If x is an integer, what is the value of x? (1) x  x^2 = 2 (2) x^2  x = 2
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Originally posted by rohitgoel15 on 01 Feb 2012, 22:58.
Last edited by Bunuel on 04 Dec 2012, 02:53, edited 1 time in total.
Edited the question.



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If x is an integer, what is the value of x? 1) x  x^2 = 2 x^2 is always positive. so x^2 = x^2
x  x^2 is negative since X^2 > x for an integer
(x  x^2) = 2 x + x^2 = 2 x^2 x  2 = 0 x^2 2x +x  2 = 0 (x2) (x+1) x = 2 or x = 1 , two values , not sufficient. 2) x2  x = 2 x^2  x is positive since X^2 > x for an integer
x can be positive or negative. so two scenarios.
x^2 x = 2 x^2 x  2 = 0 (x2)(x+1) = 0
or
x^2 + x = 2 x^2 + x  2 = 0 (x+2) (x1) = 0
Multiple values so not sufficient.
(1) + (2) , still not sufficient.



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a. squaring both sides we get x^2 + x^4  2x* x^2 = 4 meaning, x^2 (x1)^2 = 4 thus x = 2  1 not sufficient. b same process and we get x = 2 2 not sufficient. a+b gives x = 2. thus C it is.
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Originally posted by amit2k9 on 02 Feb 2012, 01:39.
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Re: value of x [#permalink]
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If x is an integer, what is the value of x? 1) x  x2 = 2 2) x2  x = 2 Hi Rohit,these type of questions are extremely easy.they just seem to be intimidating but they are not . You just need to know one concept [x]= x if x is positive [x]=x if x is negative. Now take 1) x  x2 = 2 x2 is always positive. x  x2 is negative since x^2>x x^2x=2.The value of X can be obtained as 2,1. Statement alone is not sufficient From 2) Similarly we get 2 equations x^2x=2 and x^2+x=2 depeding upon whether X is positive or negative respectively which we dont know . Statement 2 alone is not sufficient .
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rohitgoel15 wrote: If x is an integer, what is the value of x? 1) x  x2 = 2 2) x2  x = 2 I saw the solution and I think i cant even get close. On the test, I would prefer not to solve this question. But is there a short way to make an educated guess. Answer is not E as given in above posts, it's C. Also note that 1 and 1 do not satisfy statement (2). If x is an integer, what is the value of x?(1) x  x^2 = 2. First of all: \(x^2=x^2\) (as \(x^2\) is a nonnegative value). Square both sides: \((xx^2)^2=4\) > factor out \(x\): \(x^2*(1x)^2=4\) > as \(x\) is an integer then \(x=2\) or \(x=1\) (by trial and error: the product of two perfect square is 4: 1*4=4 or 4*1=4). Not sufficient. (2) x^2  x = 2 > square both sides: \((x^2x)^2=4\) > factor out \(x\): \(x^2*(x1)^2=4\) > as \(x\) is an integer then \(x=2\) or \(x=2\). Not sufficient. (1)+(2) Intersection of the values from (1) and (2) is \(x=2\). Sufficient. Answer: C. Hope it's clear.
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Re: If x is an integer, what is the value of x? [#permalink]
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03 Feb 2012, 09:46
Bunuel wrote: rohitgoel15 wrote: If x is an integer, what is the value of x? 1) x  x2 = 2 2) x2  x = 2 I saw the solution and I think i cant even get close. On the test, I would prefer not to solve this question. But is there a short way to make an educated guess. Answer is not E as given in above posts, it's C. Also note that 1 and 1 does not satisfy statement (2). If x is an integer, what is the value of x?(1) x  x^2 = 2. First of all: \(x^2=x^2\) (as \(x^2\) is a nonnegative value). Square both sides: \((xx^2)^2=4\) > factor out \(x\): \(x^2*(1x)^2=4\) > as \(x\) is an integer then \(x=2\) or \(x=1\) (by trial and error: the product of two perfect square is 4: 1*4=4 or 4*1=4). Not sufficient. (2) x2  x = 2 > square both sides: \((x^2x)^2=4\) > factor out \(x\): \(x^2*(x1)^2=4\) > as \(x\) is an integer then \(x=2\) or \(x=2\). Not sufficient. (1)+(2) Intersection of the values from (1) and (2) is \(x=2\). Sufficient. Answer: C. Hope it's clear. Thanks for the reply Bunuel. I didnt understand the factorization part in your post. It would be great if you can simplify the parts. But is there a mistake in the below post? amit2k9 wrote: a. squaring both sides we get x^2 + x^4  2x* x^2 = 4 meaning, x^2 (x1)^2 = 4 thus x = 2  1 not sufficient.
b same process and we get x = 2 2 not sufficient.
a+b gives x = 2.
thus C it is.
Now take 1) x  x2 = 2
x2 is always positive. x  x2 is negative since x^2>x
x^2x=2.The value of X can be obtained as 2,1. Statement alone is not sufficient
From 2) Similarly we get 2 equations x^2x=2 and x^2+x=2 depeding upon whether X is positive or negative respectively which we dont know . Statement 2 alone is not sufficient .



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Re: If x is an integer, what is the value of x? [#permalink]
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03 Feb 2012, 10:16
rohitgoel15 wrote: Bunuel wrote: rohitgoel15 wrote: If x is an integer, what is the value of x? 1) x  x2 = 2 2) x2  x = 2 I saw the solution and I think i cant even get close. On the test, I would prefer not to solve this question. But is there a short way to make an educated guess. Answer is not E as given in above posts, it's C. Also note that 1 and 1 does not satisfy statement (2). If x is an integer, what is the value of x?(1) x  x^2 = 2. First of all: \(x^2=x^2\) (as \(x^2\) is a nonnegative value). Square both sides: \((xx^2)^2=4\) > factor out \(x\): \(x^2*(1x)^2=4\) > as \(x\) is an integer then \(x=2\) or \(x=1\) (by trial and error: the product of two perfect square is 4: 1*4=4 or 4*1=4). Not sufficient. (2) x2  x = 2 > square both sides: \((x^2x)^2=4\) > factor out \(x\): \(x^2*(x1)^2=4\) > as \(x\) is an integer then \(x=2\) or \(x=2\). Not sufficient. (1)+(2) Intersection of the values from (1) and (2) is \(x=2\). Sufficient. Answer: C. Hope it's clear. Thanks for the reply Bunuel. I didnt understand the factorization part in your post. It would be great if you can simplify the parts. But is there a mistake in the below post? amit2k9 wrote: a. squaring both sides we get x^2 + x^4  2x* x^2 = 4 meaning, x^2 (x1)^2 = 4 thus x = 2  1 not sufficient.
b same process and we get x = 2 2 not sufficient.
a+b gives x = 2.
thus C it is.
Now take 1) x  x2 = 2
x2 is always positive. x  x2 is negative since x^2>x
x^2x=2.The value of X can be obtained as 2,1. Statement alone is not sufficient
From 2) Similarly we get 2 equations x^2x=2 and x^2+x=2 depeding upon whether X is positive or negative respectively which we dont know . Statement 2 alone is not sufficient .
First question: factoring out.\((xx^2)^2=4\) > \((x*(1x))^2=4\) > \(x^2*(1x)^2=4\); \((x^2x)^2=4\) > now, we want to factor out \(x\) (notice x^2=x*x and we are factoring out one x) > \((x*(x1))^2=4\) > \(x^2*(x1)^2=4\). Second question: other solutions.amit2k9 corrected his solution after my post so the answer there is correct. You also quote there saikarthikreddy's solution which I don't really understand as there are some parts in reasoning missing. Also it's not clear what is saikarthikreddy's answer. E? C? Please ask if anything remains unclear in my post.
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Re: If x is an integer, what is the value of x? [#permalink]
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18 Dec 2012, 21:04
Bunnel if you dont mind can you explain the factorizing part bit elaborately...Am totally not able to understand the second statement factorization
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Re: If x is an integer, what is the value of x? [#permalink]
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Re: If x is an integer, what is the value of x? [#permalink]
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Bunuel wrote: skamal7 wrote: Bunnel if you dont mind can you explain the factorizing part bit elaborately...Am totally not able to understand the second statement factorization (2) x^2  x = 2 > square both sides: \((x^2x)^2=4\). Since \(x^2=x^2\), then we have that \((x^2x)^2=4\). Factor out \(x\): \(x^2*(x1)^2=4\) > \(x^2*(x1)^2=4\). Hi Bunuel, I understood what you did, what I didnt understand is why you squared the equation before simplifying it. What I knew about Mods, my line of reasoning is similar to what Apex231 did. I was just wondering about the approach that you took, squaring and then moving forward, clearly I am missing something here.. can you please explain.



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Re: If x is an integer... [#permalink]
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09 Jan 2013, 10:48
1) \(x  x^2 = 2\) Putting x as 2 > \(2  4 = 2\) Putting x as 1 > \(1  1 = 2\) x has two values (2, 1) Hence NOT SUFFICIENT. 2) \(x^2  x = 2\) Putting x as 2 > \(4  2 = 2\) Putting x as 2 > \(4  3 = 2\) x has two values (2, 2) Hence NOT SUFFICIENT. Combining 1 & 2 gives \(x =2\). Hence (C) is the answer.PS: how did we arrive into 2, 2, 1 roots? Either pick numbers that are + or  integers or use algebraic approach below: For choice 1 consider positive absolute value: \(x  x^2 = 2\) \(x^2  x + 2 = 0\)  no possible roots for this equation. For choice 1 consider negative value: \((x  x^2) = 2\) \(x^2  x 2 = 0\) \(x = 2\) and \(x = 1\)
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Re: If x is an integer... [#permalink]
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09 Jan 2013, 10:59
sambam wrote: If x is an integer, what is the value of x?
1) x  x^2 = 2 2) x^2  x = 2 A tough one indeed. This question is not at all a sub700 level question. IMO it is atleast 730 level question. Statement 1 yields 2 cases, among which one provides nonreal numbers. The two real number values of x are 2,1. Insufficient. Statement 2 yields 4 cases as well, among which 2 provide non real numbers. The 2 real number solutions of x are (2 and 1) and (2 and 1) respectively. One doesn't satisfies the statement 2 and thus is not considered. 3 values, hence insufficient. The common value in statement 1 solution and statement 2 solution is 2. Hence x=2 . +1C
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Re: If x is an integer, what is the value of x? [#permalink]
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If x is an integer, what is the value of x?
(1) x  x^2 = 2
x  x^2 = 2 (x^2 is ALWAYS greater than or equal to zero so we can drop the absolute value sign) xx^2 = 2 Two cases, positive and negative Positive: x>1 xx^2 = 2 x^2 + x  2 = 0 x^2  x + 2 = 0 (can this be factored out?)
Negative: x<1 x + x^2 = 2 x^2  x 2 = 0 (x2) * (x+1) = 0 x=2, x=1
Here, we have two possible values for x. INSUFFICIENT
(2) x^2  x = 2
x^2  x = 2 Two cases for x, positive and negative. x>= 0 x^2  x = 2 x^2  x  2 = 0 (x2) * (x+1) = 0 x = 2, x = 1 two values, one less than zero one greater than zero. INSUFFICIENT
x<0 x^2  x =2 x^2 + x  2 = 0 (x+2) * (x1) = 0 x=2, x=1 two values for x but one is greater than zero and one is less than zero. INSUFFICIENT
See, I would say E in this case. Where did I go wrong?
Solving it another way...
(1) x  x^2 = 2 A stated above x^2 is greater than or equal to 0 so we can drop the absolute value signs.
(xx^2)^2 = 4 x=2, x=1
We get two values. INSUFFICIENT
(2) x^2  x = 2 As with #1 we can get rid of the outer absolute value signs by squaring. x^2  x = 2 (x^2  x) = 2 (x^2  x)^2 = 4 (remember, x^2 = x^2) (x^2  x)^2 = 4 x(x  1)^2 = 4 x= 2, x=2
we get two values INSUFFICIENT
1+2 we get an intersection of x=2 SUFFICIENT
Here is my question. There are many times where I have correctly used the first method (taking the positive and negative cases to solve) to solve problems and this seems like it could be one of those problems. Why is it that with this problem, that method appears to be incorrect?
Thanks!



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Re: If x is an integer, what is the value of x? [#permalink]
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07 Jul 2013, 23:26
WholeLottaLove wrote: If x is an integer, what is the value of x?
(1) x  x^2 = 2
x  x^2 = 2 (x^2 is ALWAYS greater than or equal to zero so we can drop the absolute value sign) xx^2 = 2 Two cases, positive and negative Positive: x>1 xx^2 = 2 x^2 + x  2 = 0 x^2  x + 2 = 0 (can this be factored out?)
Negative: x<1 x + x^2 = 2 x^2  x 2 = 0 (x2) * (x+1) = 0 x=2, x=1
Here, we have two possible values for x. INSUFFICIENT
(2) x^2  x = 2
x^2  x = 2 Two cases for x, positive and negative. x>= 0 x^2  x = 2 x^2  x  2 = 0 (x2) * (x+1) = 0 x = 2, x = 1 two values, one less than zero one greater than zero. INSUFFICIENT
x<0 x^2  x =2 x^2 + x  2 = 0 (x+2) * (x1) = 0 x=2, x=1 two values for x but one is greater than zero and one is less than zero. INSUFFICIENT
See, I would say E in this case. Where did I go wrong?
Thanks! This is not the best way to solve this question. Also, notice that: xx^2<0 for x<0 and x>1 and xx^2>0 for 0<x<1.
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Re: If x is an integer, what is the value of x? [#permalink]
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11 Jul 2013, 12:49
If x is an integer, what is the value of x?
(1) x  x^2 = 2
x  x^2 = 2 (we can drop the inner absolute value signs because x^2 is always >= 0) x  x^2 = 2
Normally, there would be a positive and a negative case Positive: 0<x<1 Negative: x>1, x<0
However, because x must be an integer, there is no positive case to test because only a fraction between 0 and 1 will provide a positive case.
Negative: x>1, x<0 x  x^2 = 2 (xx^2) = 2 x + x^2 = 2 x^2  x  2 = 0 (x  2)(x + 1) = 0 x=2, x=1 Both values of x satisfy their given ranges (2>1 and 1<0) So we are left with two possible correct answers INSUFFICIENT
(2) x^2  x = 2
x^2  x = 2 Two cases:
X is an integer so it must be greater than or equal to 1 or less than or equal to 1. This means that x^2  x = 2 will always be positive but x could be positive or negative.
x^2  x = 2 x>0 (x^2  x) = 2 x^2  x  2 = 0 (x  2)(x + 1) = 0 x=2, x=1 x=2 is Valid OR x<0 (x^2  (x)) = 2 x^2 + x  2 = 0 (x + 2) (x  1) = 0 x=2, x=1 x=2 is Valid
So, as with #1, we have two valid solutions for x, 2, 2
1+2) The valid solutions for #1 are 2 and 1, the valid solutions for #2 are 2 and 2. The only common number between them is 2. SUFFICIENT
Could someone show me how to factor out both sides like Bunuel did in his explanation?
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25 Aug 2014, 06:23
Hey Bunuel,
I drew up the same solution but I chose E because after combining (1) and (2) I saw it as x=2 or x=2 or x=1. Can you explain why this is wrong?



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Hey bunuel, I got x=1,2 from the first statement. But in the second statement, i took positive and negative possibilities and came up with 4 equations. the 4 equations are 1.(x^2)x=2 2.(x^2)+x=2 3. (x^2)+x=2 4. (x^2)x=2 Out of these four equations, according to me, only the 1st and the second have real solutions. So from the 1st equation above i got x=1,2(Same as the result from our First Statement ) and from the second equation i got x=1,2. But everywhere people have written that they got x=2,2 from the second statement. Can you please explain it to me as to what the issue is.



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Re: If x is an integer, what is the value of x? [#permalink]
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NikhilDev wrote: Hey bunuel, I got x=1,2 from the first statement. But in the second statement, i took positive and negative possibilities and came up with 4 equations. the 4 equations are 1.(x^2)x=2 2.(x^2)+x=2 3. (x^2)+x=2 4. (x^2)x=2 Out of these four equations, according to me, only the 1st and the second have real solutions. So from the 1st equation above i got x=1,2(Same as the result from our First Statement ) and from the second equation i got x=1,2. But everywhere people have written that they got x=2,2 from the second statement. Can you please explain it to me as to what the issue is. Does 1, or 1 satisfy the equation? No. So, x cannot be 1 or 1. Next, you get x^2  x = 2 for positive x, so when solving you should discard negative solutions. Similarly, you get x^2 + x = 2 for negative x, so when solving you should discard positive solutions.
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