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

Question

(1) |x - |x^2|| = 2 
(2) |x^2 - |x|| = 2

Bunuel · Accepted Answer

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 non-negative value). Square both sides:  (x-x^2)^2=4  --> factor out  x :  x^2*(1-x)^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^2-|x|)^2=4  --> factor out  |x| :  x^2*(|x|-1)^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.

Apex231 · Answer

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
(x-2) (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
(x-2)(x+1) = 0

or

x^2 + x = 2
x^2 + x - 2 = 0
(x+2) (x-1) = 0

Multiple values so not sufficient.

(1) + (2) , still not sufficient.

chetan2u · Answer

Looking individually at the two statements, both are clearly not sufficient as each is a quadratic equation and will have two roots.....

Combined..
 |x-|x^2||=|x^2-|x|| ...
this shows that x is +ive, otherwise  |x-|x^2||>|x^2-|x|| ...
Now lets solve any equation
 |x^2 -|x|| =2... ..............
x^2-x=2.. 
or  x^2-x-2 =0...................x^2-2x+x-2=0...................(x-2)(x+1) = 0...................... x = 2 ...or... -1 , BUT x is +ive so x=2
Suff
C

amit2k9 · Answer

a. 
squaring both sides we get 
x^2 + x^4 - 2x* x^2 = 4
meaning, x^2 (x-1)^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.

saikarthikreddy · Answer

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 if x is positive
 =-x if x is negative.

Now take 1) |x - |x2|| = 2

|x2| is always positive. |x - |x2|| is negative since x^2>x

x^2-x=2.The value of X can be obtained as 2,-1.
Statement alone is not sufficient

From 2) Similarly we get 2 equations x^2-x=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 .

rohitgoel15 · Answer

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?

Bunuel · Answer

First question: factoring out. 
 (x-x^2)^2=4  -->  (x*(1-x))^2=4  -->  x^2*(1-x)^2=4 ;
 (x^2-|x|)^2=4  --> now, we want to factor out  |x|  (notice x^2=|x|*|x| and we are factoring out one |x|) -->  (|x|*(|x|-1))^2=4  -->  x^2*(|x|-1)^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.

skamal7 · Answer

Bunnel if you dont mind can you explain the factorizing part bit elaborately...Am totally not able to understand the second statement factorization

Bunuel · Answer

(2) |x^2 - |x|| = 2 --> square both sides:  (x^2-|x|)^2=4 . Since  x^2=|x|^2 , then we have that  (|x|^2-|x|)^2=4 . Factor out  |x| :  |x|^2*(|x|-1)^2=4  -->  x^2*(|x|-1)^2=4 .

jakelong1 · Answer

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.

PrashantPonde · Answer

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

Marcab · Answer

A tough one indeed.
This question is not at all a sub-700 level question. IMO it is atleast 730 level question.

Statement 1 yields 2 cases, among which one provides non-real 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

Marcab · Answer

Please do proper search before posting.
This question has already been discussed at if-x-is-an-integer-what-is-the-value-of-x-126958.html.
Here are the rules for posting in the forum.
rules-for-posting-please-read-this-before-posting-133935.html.
Topic locked.

WholeLottaLove · Answer

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)
|x-x^2| = 2
Two cases, positive and negative
Positive: x>1
x-x^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
(x-2) * (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
(x-2) * (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) * (x-1) = 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.

(x-x^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!

Bunuel · Answer

This is not the best way to solve this question. Also, notice that: x-x^2<0 for x<0 and x>1 and x-x^2>0 for 0

WholeLottaLove · Answer

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)
|x-x^2| = 2
Two cases, positive and negative
Positive: x>1
x-x^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
(x-2) * (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
(x-2) * (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) * (x-1) = 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.

(x-x^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!

fra · Answer

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?

Bunuel · Answer

x = -1 does not satisfy either of the statements.
x = -2 does not satisfy the first statement.

ND9129 · Answer

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.

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

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

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