What is the value of x? (1) |x + 2| = 2|x - 2| (2) x > 2

(1) |x+2|=2|x−2|
Squaring both sides, we get
(x+2)^2 = 4(x-2)^2
Solving we get (3x-2)(x-6)= 0
x = 6 or 2/3
Not Sufficient

2) x>2.
Clearly not Sufficient

Combining 1 &2
we get x = 6. Hence sufficient

Answer is C

1)from four possibilities we get x=6,2,2/3...............SO INSUFFICIENT
2) Clearly insufficient
From both we get x=6

OA:C

#1
|x+2|=2|x−2|
possible value of x = 6,2/3
insufficient
#2
x>2
insufficient
from 1&2
x=6
sufficient
IMO C


What is the value of x?

(1) |x+2|=2|x−2|
(2) x>2

positive:|x+2|≥0…x≥-2…negative:x<-2
positive:|x-2|≥0…x≥2…negative:x<2
range:__--__(-2)__-+___(2)__++__

(1) |x+2|=2|x−2| insufic

\(x>2:(x+2)=2(x-2)…x=6…(x≥2=valid)\)
\(-2≤x<2:-(x+2)=2(x-2)…-x-2=2x-4…x=2/3…(-2≤x<2=valid)\)
\(x<-2:-(x+2)=-2(x-2)…-x-2=-2x+4…x=6…(x<-2=invalid)\)

\(x=(6,2/3)\)

(2) x>2 insufic

(1&2) sufic
\(x>2:(x+2)=2(x-2)…x=6…(x≥2=valid)\)

Ans (C)

We are to determine the value of x.

Statement 1: |x+2|=2|x-2|
squaring both sides yield (x+2)^2=4*[(x-2)^2]
x^2 + 4x + 4 = 4x^2 - 16x +16
3x^2 - 20x + 12 = 0
(3x-2)(x-6)=0
x=2/3 and x=6
When x=2/3, |2/3+2| = 2|2/3-2|
8/3=2|-4/3|
8/3=8/3
RHS=LHS, hence x=2/3 is a root of the equation
when x=6, |6+2|=2|6-2|
8=8, LHS=RHS, hence x=6 is a root of the equation.

Now we have two values of x satisfying the given equation. x can be 2/3 and x can be 6. We are unable to narrow down to a unique value for x, hence statement 1 is insufficient on its own.

Statement 2: x>2
Clearly insufficient because x can be 3, 4, 5, etc. Statement 2 does not lead to a unique value of x hence it is not sufficient on its own.

1+2:
we know from 1 that x can be 3/2, or x can be 6. Statement 2 says that x>2, hence we can categorically say that x=6.

Combining both statements is, therefore, sufficient.

C is the answer.

What is the value of x?

(Statement1) |x +2|= 2|x—2|
—> square both sides:
\((x+2)^{2} = (2(x—2))^{2}\)

—>( x+2 —2x+4)(x+2 +2x—4)= 0
(—x+6)(3x—2)= 0
x= 6 and \(x= \frac{2}{3}\)
Insufficient

(Statement2) x >2
Clearly insufficient

Taken together 1&2,
x= 6, \(x= \frac{2}{3}\) and x >2
—> x=6
Sufficient

The answer is C

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|x+2| = 2|x-2|
will be broken into 4 cases if solved algebraically -

case 1

|x+2| = x+2 if x>-2

With this inequality, we have 2 sub conditions:

case 1a : |x-2| = x-2 if x>2

this makes the equation as

x+2 = 2(x-2)
x=6

here x>-2 and x>2 makes x=6 a valid answer hence we keep it.


case 1b: |x-2| = -x+2 if x<2

this makes the equation as

x+2 = 2(-x+2)
x=2/3

here x>-2 and x<2 makes x=2/3 a valid answer hence we keep it.


Case 2 - |x+2| = -x-2 if x<-2

This will again have 2 cases

case 2a : |x-2| = x-2 if x>2

Since x>2 and x<-2 so this is invalid case.

Case 2b: |x-2| = -x+2 if x< 2

-x -2 = 2(2-x)
x=6 which does not fit in x<-2 and x<2 hence invalid

So we have 2 answers now. x =6 and x = 2/3
so option b would also be required to get x=6 as the answer.

Hence the answer should be "C" both options required.

X=?
Statement 1= |x+2|/|X-2|=2

If we remove the modulus we get x=-2 or x=2. Hence in sufficient

Statement 2 states that x<2

Still insufficient it could be 3,4 etc

When combined and values added only x=6 fits

IMO C

Posted from my mobile device

Option C
By solving the stmt 1 we get three values of x =6,2,2/3
insuff
stmt 2 is clearly insuff

1+2 indicates x can only hav the value 6

Posted from my mobile device

This is a value kind of DS question where we need to find a unique value for x.

Performing a visual scan of the statements, it’s very clear that statement 2 alone will not help us find a unique value for x, since x>2 represents a range. Therefore, the first thing I would do on this question is I would eliminate options B and D since they cannot be the answers. At this stage, I am left with options A, C and E as my possible answers.

Whenever such an opportunity presents itself, where you think some options can be eliminated, grab it with both hands. This means that you will have more time to spend on the statements and options that matter.

Remember that |x| = √\(x^2\).

From statement 1 alone, |x+2| = 2|x-2|. The absolute value terms can be re-written based on the concept discussed above.

So, |x+2| = √\((x+2)^2\) and |x-2| = √\((x-2)^2\). Substituting this in the equation, we get,

√\((x+2)^2\)= 2√\((x-2)^2\). Squaring both sides, we have,

\((x+2)^2\) = 4\((x-2)^2\).
Expanding the expressions based on identities and solving for the values, we obtain two values for x viz x = \(\frac{2}{3}\) and x = 6. But, we need a unique value for x. Hence, statement 1 alone is insufficient.
Answer option A can be eliminated. Possible answer options are C or E.

Combining the data from statements 1 and 2, we can say that x has to be 6 since that is the only value of x that satisfies the conditions given in both the statements.

The combination of statements is sufficient. Answer option E can be eliminated.
The correct answer option is E.

In absolute value questions where you see a modulus on either side of an equation, it’s a good idea to use the |x| = √\(x^2\) concept. This will help you to reduce the equation to a quadratic which can then be solved for the values of x.
Also remember to not jump to a conclusion that a modulus will give you two values for x and hence insufficient. That is not the case in all questions. Only when you analyse will you find out.

Hope that helps!