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GCT Absolute Value 2

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by richachampion » Tue Nov 29, 2016 3:16 am
What is the range of all the roots of |X²-2| = X

A. 4
B. 3
C. 2
D. 1
E. 0
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by richachampion » Tue Nov 29, 2016 3:16 am
OA: D
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by GMATGuruNY » Tue Nov 29, 2016 6:08 am
richachampion wrote:What is the range of all the roots of |X²-2| = X

A. 4
B. 3
C. 2
D. 1
E. 0
Since the absolute value on the left side cannot be equal to a negative value, the right side of the equation must be NONNEGATIVE.
Thus, only nonnegative values for x are viable here.

Case 1: x²-2 = x
x² - x - 2 = 0
(x-2)(x+1) = 0.
x=2 or x=-1.
Since x must be nonnegative, only x=2 is viable.

Case 2: x²-2 = -x
x² + x - 2 = 0
(x+2)(x-1) = 0.
x=-2 or x=1.
Since x must be nonnegative, only x=1 is viable.

The range of the two roots = greater root - smaller root = 2-1 = 1.

The correct answer is D.
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by Brent@GMATPrepNow » Tue Nov 29, 2016 6:08 am
richachampion wrote:What is the range of all the roots of |x² - 2| = x

A. 4
B. 3
C. 2
D. 1
E. 0
When solving equations involving ABSOLUTE VALUE, there are 3 steps:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

EITHER x² - 2 = x OR x² - 2 = -x

x² - 2 = x
Rearrange: x² - x - 2 = 0
Factor: (x - 2)(x + 1) = 0
Solve: x = 2 or x = -1

x² - 2 = -x
Rearrange: x² + x - 2 = 0
Factor: (x + 2)(x - 1) = 0
Solve: x = -2 or x = 1

Plug solutions into original equation to check for extraneous roots....
Plug in x = 2 to get |2² - 2| = 2
Evaluate: |2| = 2 WORKS!
So, x = 2 is a solution

Plug in x = -1 to get |(-1)² - 2| = -1
Evaluate: |-1| = -1 DOESN'T WORK
So, x = -1 is a NOT solution

Plug in x = -2 to get |(-2)² - 2| = -2
Evaluate: |2| = -2 DOESN'T WORK
So, x = -2 is a NOT solution

Plug solutions into original equation to check for extraneous roots....
Plug in x = 1 to get |1² - 2| = 1
Evaluate: |-1| = 1 WORKS!
So, x = 1 is a solution

So, the only two valid solutions are x = 2 and x = 1
So, the range is 1
Answer: D

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- Solving equations with absolute value: https://www.gmatprepnow.com/module/gmat ... /video/972
- Solving quadratic equations: https://www.gmatprepnow.com/module/gmat ... /video/964
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by [email protected] » Tue Nov 29, 2016 10:52 am
Hi richachampion,

This question can be solved rather easily with some 'brute force' math. The answer choices are all small integers, so we know that the range can't be very big. Chances are good that we can just TEST VALUES until we find the range.

With |X^2 - 2| = X we know that X cannot be negative (since the result of an absolute value calculation can't be negative). So let's get to work TESTing small integers - we really just have to take notes and look for a pattern.

IF...
X = 0, we get |-2| = 0, which is NOT correct.
X = 1, we get |-1| = 1, so X COULD be 1
X = 2, we get |-2| = 2, so X COULD be 2
X = 3, we get |7| = 3, which is NOT correct.

If we increase the value of X, then we'll just end up with an absolute value that is farther away from the value of X, so we can stop working here. For the sake of thoroughness, you might consider TESTing non-integers (for example, X = 1/2), but you'll rather quickly see that the equation won't balance out (the denominators will be different)...

IF...
X = 1/2, we get |-7/4| = 1/2, which is NOT correct.

Thus, the only solutions are 1 and 2, so the range is 1.

Final Answer: D

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by Matt@VeritasPrep » Thu Dec 08, 2016 9:11 pm
|x² - 2| = x

We know that |z| is always = √z², so let's write it that way:

√(x² - 2)² = x

then square both sides:

(x² - 2)² = x²

then expand:

x� - 4x² + 4 = x²

then solve:

x� - 5x² + 4 = 0

(x² - 4) * (x² - 1) = 0

Which gives us

(x² - 4) = 0 or (x² - 1) = 0

and we know that |x² - 2| gives a positive result, so x must be positive.

That means (x² - 4) only has the solution x = 2, and (x² - 1) = 0 only has the solution x = 1.

That gives us our two solutions, and we're done!
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by GMATsid2016 » Fri Dec 09, 2016 8:37 am
Since the absolute value on the left side cannot be equal to a negative value, the right side of the equation must be NONNEGATIVE.
Thus, only nonnegative values for x are viable here.
Sir, can you please explain above with example?

Thanks,

Sid
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by [email protected] » Fri Dec 09, 2016 11:27 am
Hi Sid,

Absolute value calculations turn negative results into positive ones.

For example, if we're dealing with |X - 3|...

IF....
X = -1, then | -1 - 3| = |-4| = 4
X = 2, then |2 - 3| = |-1| = 1
X = 3, then |3 - 3| = |0| = 0
X = 4, then |4 - 3| = |1| = 1

Using that knowledge, we know that |X^2 - 2| can NEVER end in a negative result (it could end in a 0 result, but that would be the minimum possible result).

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