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Re: If x is an integer, what is the value of x? [#permalink]
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04 Feb 2016, 06:05
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) 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. if x^2 = 4 then isn't x= (2,2) ?



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Re: If x is an integer, what is the value of x? [#permalink]
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04 Feb 2016, 06:10
email2vm 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) 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. if x^2 = 4 then isn't x= (2,2) ? Yes, but if x=2, then x^2*(1x)^2 does not equal to 4.
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Re: If x is an integer, what is the value of x? [#permalink]
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04 Feb 2016, 06:52
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. You are making a mistake while interpreting different scenarios after you open up the absolute sign. Per statement 2, x^2x=2 Consider 2 cases, Case 1: when \(x \geq 0\), \(x =x\) >\(x^2x=2\) becomes \(x^2x=2\) > \(x^2x=\pm 2\) , giving you 2 quadratic equations \(x^2x2=0\) and \(x^2x+2=0\) (no real solutions, eliminate). From \(x^2x2=0\) , you get x=2 and x=1 but x=1 is NOT allowed as you are assuming that for this case, \(x \geq 0\). Thus x=2 is the only possible solution from this scenario. Case 2: when \(x<0\)> \(x = x\) > \(x^2x=2\) becomes \(x^2+x=2\)> \(x^2+x=\pm 2\) , giving you 2 quadratic equations again, \(x^2+x2=0\) and\(x^2+x+2=0\) (no real solutions, eliminate). From \(x^2+x2=0\), you get x=2 and x=1 but x=1 is NOT allowed as you are assuming that for this case, \(x < 0\). Thus x=2 is the only possible solution from this scenario. Finally, when you combine both the scenarios, you see that the allowed values are \(x=\pm 2\). Thus this statement is NOT sufficient to answer the question asked. Hope this help. You have to be very careful about your assumptions in absolute value questions.



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Re: If x is an integer, what is the value of x? [#permalink]
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04 Feb 2016, 08:46
rohitgoel15 wrote: If x is an integer, what is the value of x?
(1) x  x^2 = 2 (2) x^2  x = 2 Question : x=?Statement 1: x  x^2 = 2 The solutions of the given equation are x=2 and 1 NOT SUFFICIENT Statement 2: x^2  x = 2The solutions of the given equation are x=2 and 2 NOT SUFFICIENT Combining the two statementsx = 2 SUFFICIENT Answer: Option C
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Re: If x is an integer, what is the value of x? [#permalink]
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04 Feb 2016, 17:31
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If x is an integer, what is the value of x? (1) x  x^2 = 2 (2) x^2  x = 2 In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), it becomes x^2=x^2 > xx^2=2, +2. In xx^2=2, +2, x^2x2=0, (x2)(x+1)=0 > x=2, 1, which is not unique and not sufficient. For 2), in x^2x=2, 2, x^2=x^2 is derived. x^2x=2, x^2x2=0, (x2)(x+1)=0 > x=2, 1. 1 is impossible. x=2 > x=2,2, which is not unique and not sufficient. When 1) & 2), x=2 is unique, which is sufficient. Therefore, the answer is C. > For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x is an integer, what is the value of x? [#permalink]
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07 May 2016, 09:47
bimalr9 wrote: If x is an integer, what is the value of x?
1. xx^2 = 2 2. x^2 x =2 Looking individually at the two statements, both are clearly not sufficient as each is a quadratic equation and will have two roots..... Combined.. \(xx^2=x^2x\)... this shows that x is +ive, otherwise \(xx^2>x^2x\)... Now lets solve any equation \(x^2 x =2... .............. x^2x=2..\) or \(x^2x2 =0...................x^22x+x2=0...................(x2)(x+1) = 0...................... x = 2 ...or... 1\), BUT x is +ive so x=2 Suff C
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Re: If x is an integer, what is the value of x? [#permalink]
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07 May 2016, 10:43
A slightly different approach:
Statement 1 X is 1 or 2 Insufficient
Statement 2 X is 2 or 2 Insufficient
Combined 2 is the only common root for both equations = hence X must be 2. Answer = C



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Re: If x is an integer, what is the value of x? [#permalink]
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30 Sep 2017, 09:26
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) 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. Hi How to approach the factorisation without hit and trial? Any ideas?




Re: If x is an integer, what is the value of x?
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