If |x^2 − 12| = x, which of the following could be the value of x?

Question

A. –4 
B. –3 
C. 1 
D. 2 
E. 3

chetan2u · Accepted Answer

Hi,
In MODULUS, always be careful of EQUATING two sides..

Here ONE important point is that LHS is positive, so RHS will also be positive, SO x will be +IVE..
A and B are eliminated..
Now try to substitute remaining choices
C. 1 
 |x^2 − 12| = x ..
 |1^2-12|=1………. 11=1 .. NO

D. 2 
 |x^2 − 12| = x ..
 |2^2-12|=2…….. 7=2 .. NO

E. 3
 |x^2 − 12| = x ..
 |3^2-12|=3…………. 3=3 .. YES

ans  E

Youraisemeup · Answer

well, a mod here indicates that x is positive.
Eliminate a and b immediately.
 Now, you may substitute values themselves..none of them except E fits the bill.

Sindbad · Answer

Another easy way and time-saver will be to substitute the values directly in the equation.
so 
A: |(-4)^2-12|=|4| ne -4 => FALSE
B: |(-3)^2-12|=|-3| ne -3 => FALSE
C: |(1)^2-12|=|-11| ne -1 => FALSE
D: |(2)^2-12|=|-8| ne 2=> FALSE
E: |(3)^2-12|=|-3| eq 3 => TRUE

Abhishek009 · Answer

Substituting the value works in most of the Problems of GMAT and is a sureshot way...

Same opinion , answer will be (E)

LogicGuru1 · Answer

|x^2 − 12| = x 
 |x^2 − 12| cannot be negative; anything that comes out of mod is always positive
so  |x^2 − 12|>0

Out of our options A= -4 B= -3 cannot be the right answer because none of them are greater than 0

Now only Option C=1 or D=2 or E=3 are contenders for right answers.

A quick Cursory crude calculation will tell you that E is the right choice 
 |-3^2-12|=3

|9-12|=3

|-3|=3

3=3

ANSWER IS E

AkshdeepS · Answer

Hi chetan2u,

Though I know the substitution method, but instinctively I went to solve the the equations given.

Can you help me how it can be done with critical value method. I am stuck up.

AkshdeepS · Answer

Can you tell me the algebraic way?

I am stuck up a bit.

QZ

ScottTargetTestPrep · Answer

Since we see that the left side of the equation must be positive, the right side also must be positive.

Thus, A and B cannot be correct since the absolute value of any quantity can’t be negative. Let’s substitute in the remaining answer choices:

|1^2 - 12 | = 1 ?

We see this is not true.

|2^2 - 12| = 2 ?

We see this is not true

|3^3 - 12| = 3 ?

We see this is true.

Answer: E

CesarAMSTech · Answer

I'll help. First of all, x is the outcome of an absolute value so x=>0 Then we have to analyze the two possible scenarios of what we have inside the absolute value so that we can eliminate the absolute value: 1. X^2-12=>0, x^2=>12 So that x=<-sqrt(12) or x=>sqrt(12) The only possible scenario according to the constraint that x can take any non negative value is the second one so we can say that x has to be greater than or equal to approximately 3.4 Then we would have the quadratic equation x^2-x-12=0 with roots x=-3 or x=4 we eliminate the first one because x can not take negative values. 2. X^2-12<0, x^2<12 So that -sqrt(12)

AkshdeepS · Answer

Thanks Cesar to clarify.

Posted from my mobile device

AkshdeepS · Answer

The algebraic way:

|x^2 − 12| = x

Case 1:

x^2 - 12 = x

x^2 - x = 12

x (x - 1) = 12

We need two consecutive numbers whose products is 12.

So 4(4-1) = 12

4*3 = 12

But we don't have "4" in the answer choices.

Case 2:

x^2 - 12 = -x

x^2 + x = 12

x (x + 1) = 12

We need two consecutive numbers whose products is 12.

3 (3 + 1) = 12

3*4 = 12

12 = 12

So x = 3

(E)

AvinashT2303 · Answer

Hello, is there any algebraic way to solve this? can u plz explain

Posted from my mobile device

AkshdeepS · Answer

HERE IT IS: https://gmatclub.com/forum/if-x-2-12-x- ... l#p2052961 Please read the complete thread before raising queries. Hope this helps. QZ

luisdicampo · Answer

Deconstructing the Question

We are given:

|x^2 - 12| = x

Since the left side is an absolute value, it is always nonnegative. Therefore:

x ≥ 0

So we can immediately eliminate negative choices.

Now solve by splitting into cases.

Step-by-step

Case 1:  x^2 - 12 ≥ 0

|x^2 - 12| = x^2 - 12

Equation:

x^2 - 12 = x

x^2 - x - 12 = 0

(x - 4)(x + 3) = 0

x = 4  or  x = -3

Since  x ≥ 0 , only  x = 4  is valid, but it is not among the choices.

Case 2:  x^2 - 12 < 0

|x^2 - 12| = 12 - x^2

Equation:

12 - x^2 = x

x^2 + x - 12 = 0

(x + 4)(x - 3) = 0

x = -4  or  x = 3

Since  x ≥ 0 , only  x = 3  is valid.

Answer E

If |x^2 − 12| = x, which of the following could be the value of x?

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If |x^2 − 12| = x, which of the following could be the value of x?

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