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

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
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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.

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

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

Same opinion , answer will be (E)
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\(|x^2 − 12| = x\)
\(|x^2 − 12|\)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

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.
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Can you tell me the algebraic way?

I am stuck up a bit.

QZ
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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
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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)<x<sqrt(12)
Then according to the initial constraint 0<x<3.4

Then we would have the quadratic equation x^2+x-12=0 with roots x=3 or x=-4
We eliminate the second one because x can not take negative values.

So from the first scenario x=4 and from the second one x=3, those are the two solutions for the absolute value equation.

I hope it is clear.

Regards

César Orihuela

Posted from my mobile device

Thanks Cesar to clarify.

Posted from my mobile device
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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)
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Hello, is there any algebraic way to solve this? can u plz explain

Posted from my mobile device

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
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