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by Jay@ManhattanReview » Sat Aug 05, 2017 12:50 am
Mo2men wrote:If x is an integer, what is the value of x?

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

OA: C
Source: GMAT Pill
Given that x is an integer.

We have to get the value of x.

Statement 1: |x - |x^2|| = 2

So we have to deal with two modulus here.

One of the approaches to deal this is by applying brute force.

Remember that 0, +1, -1, +2, and -2 are must check values if you do a hit and trial.

Let's test |x - |x^2|| = 2 at these values.

1. @x = 0, |x - |x^2|| = 0 - 0 ≠ 2. Not an appropriate value!
2. @x = 1, |x - |x^2|| = 1 - 1 = 0 ≠ 2. Not an appropriate value!
3. @x = -1, |x - |x^2|| = |-1 - |-1^2|| = |-1 - 1| = 2. An appropriate value; x = -1
4. @x = 2, |x - |x^2|| = |2 - |2^2|| = |2 - 4| = 2. An appropriate value; x = 2

So we get either x = -1 or 2; no unique value. Insufficient.

Statement 2: |x^2 - |x|| = 2

Since there is no need to check the value of |x^2 - |x|| at x= 0, let's test |x^2 - |x|| at other values of x.

1. @x = 1, |x^2 - |x|| = |1^2 - |1|| = 0 ≠ 2. Not an appropriate value!
2. @x = -1, |x^2 - |x|| = |-1^2 - |-|| = 0 ≠ 2. Not an appropriate value!
3. @x = 2, |x^2 - |x|| = |2^2 - |2|| = |4 - 2| = 2. An appropriate value; x = 2
4. @x = -2, |x^2 - |x|| = |-2^2 - |-2|| = |4 - 2| = 2. An appropriate value; x = -2

So we get either x = 2 or -2; no unique value. Insufficient.

Statement 1 & 2:

From (1), we get x = -1 or 2 and from (2), we get x = 2 or -2 , thus x = -2. Sufficient!

The correct answer: C

Hope this helps!

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value of x

by GMATGuruNY » Sat Aug 05, 2017 3:08 am
If x is an integer, what is the value of x?

1)|x-|x^2||=2
2)|x^2 -|x||=2
In statement 1, |x²| is redundant: since x² cannot be negative, |x²| = x².

Statement 1: |x-x²|=2
x - x² = ±2
x(1-x) = ±2.

The resulting equation implies that x is a factor of ±2, yielding the following options:
x=±1 or x=±2.
Check which of these values are valid solutions for |x-x²| = 2.

If x=1, then |x-x²| = |1 - 1²| = 0.
If x=-1, then |x-x²| = |-1 - (-1)²| = 2.
If x=2, then |x-x²| = |2 - 2²| = 2.
If x=-2, then |x-x²| = |-2 - (-2)²| = 6.

Since it's possible that x=-1 or that x=2, INSUFFICIENT.

Statement 2: |x² -|x||=2
x²-|x| = ±2
Since x² = |x|*|x|, we can factor out |x|:
|x| (|x|-1) = ±2.

The resulting equation implies that |x| is a factor of ±2, yielding the following options:
|x|=1 or |x|=2, with the result that x=±1 or x=±2.
Check which of these values are valid solutions for |x² -|x||=2

If x=-1, then |x² -|x|| = |(-1)² - |-1|| = 0.
If x=1, then |x² -|x|| = |1² - |1|| = 0.
If x=2, then |x² -|x|| = |2² - |2|| = 2.
If x=-2, then |x² -|x|| = |(-2)² - |-2|| = 2.

Since it's possible that x=2 or that x=-2, INSUFFICIENT.

Statements 1 and 2 combined:
Both statements are satisfied only by x=2.
SUFFICIENT.

The correct answer is C.
Last edited by GMATGuruNY on Sun Jul 07, 2019 4:14 am, edited 1 time in total.
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by Mo2men » Sat Aug 05, 2017 5:13 am
GMATGuruNY wrote:
In statement 1, |x²| is redundant: since x² cannot be negative, |x²| = x².

Statement 1: |x-x²|=2
x - x² = ±2
x(1-x) = ±2.

Since x must be an integer, x=±1 or x=±2.
Check which of these values are valid solutions for |x-x²| = 2.

If x=1, then |x-x²| = |1 - 1²| = 0.
If x=-1, then |x-x²| = |-1 - (-1)²| = 2.
If x=2, then |x-x²| = |2 - 2²| = 2.
If x=-2, then |x-x²| = |-2 - (-2)²| = 6.

Since it's possible that x=-1 or that x=2, INSUFFICIENT.


Dear Mitch,

I have GENERAL question and I use Statement 1 as an example

In Absolute question with '=' sign, can I square both sides and check solutions in the ORIGINAL equation?

Thanks
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by GMATGuruNY » Mon Aug 07, 2017 2:53 am
Mo2men wrote:Dear Mitch,

I have GENERAL question and I use Statement 1 as an example

In Absolute question with '=' sign, can I square both sides and check solutions in the ORIGINAL equation?

Thanks


Yes, we can square an equation with absolute value.
But if it is possible for one side of the equation to yield a negative result, we must confirm the validity of the yielded solution(s).

|x+1| = x-2.
Here, if x<2, then the right side will yield a negative -- and thus invalid -- result.
Squaring both sides, we get:
|x+1|² = (x-2)²
x² + 2x + 1 = x² - 4x + 4
6x = = 3
x = 1/2.
The resulting solution is less than 2 and thus is invalid.

One way to confirm the validity of a resulting solution is to plug it back into the original equation.
Plugging x=1/2 into |x+1| = x-2, we get:
|1/2 + 1| = 1/2 - 2
3/2 = -3/2.
Doesn't work.
Thus, x=1/2 is NOT a valid solution for |x+1| = x-2.

The statements in blue indicate that |x+1| = x-2 has no real solution.

In the problem above:
|x-x²|=2
Here, neither side of the equation can yield a negative result.
Thus, if we solve by squaring the equation, the yielded solutions are guaranteed to be valid.
There is no reason to plug them back into the original equation.
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by Mo2men » Tue Aug 22, 2017 2:11 am
GMATGuruNY wrote: Yes, we can square an equation with absolute value.
Dear Mitch,

Is the above rule applied to inequality in general? You have posted once if both are positive. For example I'm no sure if I can square the following inequality?
|1 - x| < 1

When is WRONG to square both sides of inequality? can just cite a simple example.

Thanks in advance for your keen support
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700 Absolute Problem

by GMATGuruNY » Tue Aug 22, 2017 4:34 am
Mo2men wrote:When is WRONG to square both sides of inequality? can just cite a simple example.

Thanks in advance for your keen support
If either side of an inequality can be less than 0, I would avoid squaring the inequality.
x < y does not necessarily imply that x² < y².
If x=-1 and y=1, then x < y, but x² = y².
can square the following inequality?
|1 - x| < 1
Here, both sides are nonnegative, so we can safely square the inequality.
That said, there are more efficient ways to solve.
Last edited by GMATGuruNY on Tue Aug 22, 2017 8:15 am, edited 1 time in total.
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by Mo2men » Tue Aug 22, 2017 4:47 am
GMATGuruNY wrote:
can square the following inequality?
|1 - x| < 1
Here, both sides are nonnegative, so we can safely square the inequality.
That said, there are more efficient ways to solve.
When you mention efficient way, I would find the critical points 0 & 2 that makes both sides equal and plot them on number line and discover the ranges that hold true.

Is that more efficient the squaring both sides?

Thanks in advance for advice
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by GMATGuruNY » Tue Aug 22, 2017 5:07 am
Mo2men wrote:
GMATGuruNY wrote:
can square the following inequality?
|1 - x| < 1
Here, both sides are nonnegative, so we can safely square the inequality.
That said, there are more efficient ways to solve.
When you mention efficient way, I would find the critical points 0 & 2 that makes both sides equal and plot them on number line and discover the ranges that hold true.

Is that more efficient the squaring both sides?

Thanks in advance for advice
Your approach is fine.

Alternate approach 1:
|a-b| = the distance between a and b.
Thus, |1 - x| < 1 implies that the distance between 1 and x is less than 1.
In other words, x must be less than ONE PLACE AWAY FROM 1, as follows:
0<---1--->2
To be less than one place away from 1, x must be within the blue range above.
Thus, 0<x<2.

Alternate approach 2:
Case 1: Signs unchanged
1-x < 1
0 < x
x > 0.

Case 2: Signs changed in the absolute value
-1+x < 1
x < 2.

Thus:
0<x<2.
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