What is the sum of the different values of x such that |x+2| = 2|x-2|?

Question

What is the sum of the different values of x such that |x+2| = 2|x-2| ? A. 4 B. \frac{16}{3} C. 6 D. \frac{20}{3} E. 8 2023-12-04_21-00-05.png

Bunuel · Accepted Answer

The key point is that when expanding the absolute values, the pair |x + 2| and |x − 2| can be expanded with either the same sign (where '+' and '+' or '−' and '−' will yield the same expression) or with opposite signs (where '+' and '−' will result in the same expression as '−' and '+').

Thus, |x + 2| = 2|x - 2| can lead to either x + 2 = 2(x - 2) when expanding with the same sign, or x + 2 = -2(x - 2) when expanding with opposite signs.

x + 2 = 2(x - 2) yields x = 6. Substituting x = 6 back into |x + 2| = 2|x - 2| satisfies the equation, confirming this value as valid.
x + 2 = -2(x - 2) yields x = 2/3. Substituting x = 2/3 back into |x + 2| = 2|x - 2| also satisfies the equation, confirming this value as well.

The sum of these values, 6 + 2/3 = 18/3 + 2/3 = 20/3.

Answer: D.

rakman123 · Answer

Simple Method;

Must note; when you come across an  absolute value question  like this, always remember that you must test the value, in this case 'x', for both a positive side and a negative side. Here's what I mean;

Positive: |x+2|=2|x−2| ---> x+2=2(x-2) ---> x+2=2x-4 ---->  x=6  , this is the value for the positive.
Negative: |x+2|=2|x−2| ---> -x-2=2(x-2) ---> -x-2=2x-4 ---> 3x=2 --->  x=2/3  , this is the value for the negative.

Sum  6+2/3= 20/3.

Voila.

gmatophobia · Answer

|x+2| = 2|x-2|

Squaring both sides of the equation

(x+2)^2 = 4(x-2)^2

x^2 + 4 + 4x = 4(x^2 + 4 - 4x)

x^2 + 4 + 4x = 4x^2 + 16 - 16x

3x^2 - 20x + 12 = 0

In a quadratic equation represented by  ax^2 + bx + c , the sum of roots is given by  \frac{-b}{a}

Sum of roots =  \frac{20}{3}

Option D

Adrian_D · Answer

---------
^ Why did you only test a positive and negative for the left side of the equation, given there is an absolute value sign on both sides?

SergejK · Answer

But as x<-2 and hence x<2 would result in x=6 we can discard this answer choice as it is clearly not less than -2, correct? Additionally as C4 of x<-2 and x>= 2 together doen't make sense, we are left with 2 cases: C1: x>=2, h>=-2 and C2: x>=-2 and x<2. Is my understanding correct?

pisker · Answer

Why do you not test the second equation with a negative? Shouldn't this result in 4 scenarios?

Bunuel · Answer

Your doubt is addressed here: https://gmatclub.com/forum/what-is-the- ... l#p3310573 Hope this helps.

pisker · Answer

Thanks, I read that but that isn't quite the problem I'm having trouble understanding. Why are we not testing these 4 scenarios since we have 2 different absolute value equations:

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

pisker · Answer

Nevermind - I see it now thanks!

Gaurav07 · Answer

Hey, can you explain. Didn't quite understadn why didn't we add up all the 4 values

Venu01298 · Answer

I think its because you end up with the same 2 values as solutions for the additional two scenarios

|x+2| = 2|x-2| - original

assuming LHS positive RHS negative

x+2 = -2(x-2)

assuming LHS negative RHS positive

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

as you can see both are the same, so you would get same solutions for x

Similarly both negatives and both positives yield the same solution

BrushMyQuant · Answer

We need to find what is the sum of the different values of x such that |x+2| = 2|x-2|

As we have two absolute values so we will have three cases and he number line will be divided into three points by equating following equations to 0

x + 2 =0 and x - 2 = 0
 => x = 2 and x = -2

Number line is divided into three parts x ≤ 3, 3 ≤ x ≤ 20, x ≥ 20   
 
   -   Case 1:  x  ≤   -2
   
Now when x  ≤  - 2 then both x + 2 and x - 2 will be non-positive
 => | x+2| = -(x+2) and |x-2| = -(x-2)
=> -(x+2) = 2*-(x-2)
=> x + 2 = 2*(x-2)
=> x + 2 = 2x - 4
=> x = 6

But the range was x  ≤     -2 and 6 is NOT ≤ -2  
 =>   NO SOLUTION   in this case 
  -   Case 2:   -2 ≤ x ≤ 2
   
Now when  -2 ≤ x ≤ 2   then x + 2 will be non-negative and x - 2 will be non-positive
  => | x+2| = x+2 and |x-2| = -(x-2)
=> x+2 = 2*-(x-2)
=> x + 2 = -2x + 4
=> 3x = 2
=> x = 2/3 ~ 0.67

But the range was  -2 ≤ x ≤ 2     and -2 ≤ 0.67 ≤ 2   
  => x = 2/3 is a   SOLUTION   
   Case 3:     x ≥ 2
   
 Now when x  ≥ 2   then both x + 2 and x - 2 will be non-negative
  => | x+2| = x+2 and |x-2| = x-2
=> x+2 = 2*(x-2)
=> x + 2 = 2*(x-2)
=> x + 2 = 2x - 4
=> x = 6

But the range was   x ≥ 2 and 6 ≥ 2   
   => x = 6 is a   SOLUTION

=> Sum of different values of x = 2/3 + 6 = 20/3

So,  Answer will be D 
Hope it helps!

Watch the following video to MASTER Inequality + Absolute value Problems

https://www.youtube.com/watch?v=236auvH8agc

KarishmaB · Answer

When both sides of the = sign are surrounded by absolute values, the easiest approach usually is to square both sides. (even if it is an inequality because both sides are non negative) That method is shown by gmatophobia above so not repeating it. Those inclined to following visual approach of "distance on number line", note that |x+2| = 2|x-2| means "distance from -2 is equal to twice the distance from 2" Screenshot 2024-10-19 at 7.54.03 PM.png Distance between -2 and 2 is 4 which is split in the ratio 2:1 which means that x is at 2 - 4/3 = 2/3 Distance between -2 and 2 is 4 which is doubled to give another point where the ratio of distances is 2:1 which means x is at 2 + 4 = 6 Hence x takes 2 values: 6 and 2/3 and their sum is 20/3 Answer (D) Concept of absolute values as distance is discussed here: https://youtu.be/oqVfKQBcnrs

anushree01 · Answer

Tip  : Whenever you see  |a|=|b|  type Question :  SOLVE it by Squaring both sides

What is the sum of the different values of x such that |x+2| = 2|x-2|?

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Number line is divided into three parts x ≤ 3, 3 ≤ x ≤ 20, x ≥ 20
-Case 1: x ≤ -2 Now when x ≤ -2 then both x + 2 and x - 2 will be non-positive => \| x+2\| = -(x+2) and \|x-2\| = -(x-2) => -(x+2) = 2-(x-2) => x + 2 = 2(x-2) => x + 2 = 2x - 4 => x = 6 But the range was x ≤ -2 and 6 is NOT ≤ -2 => NO SOLUTION in this case	-Case 2: -2 ≤ x ≤ 2 Now when -2 ≤ x ≤ 2 then x + 2 will be non-negative and x - 2 will be non-positive => \| x+2\| = x+2 and \|x-2\| = -(x-2) => x+2 = 2*-(x-2) => x + 2 = -2x + 4 => 3x = 2 => x = 2/3 ~ 0.67 But the range was -2 ≤ x ≤ 2 and -2 ≤ 0.67 ≤ 2 => x = 2/3 is a SOLUTION	Case 3: x ≥ 2 Now when x ≥ 2 then both x + 2 and x - 2 will be non-negative => \| x+2\| = x+2 and \|x-2\| = x-2 => x+2 = 2(x-2) => x + 2 = 2(x-2) => x + 2 = 2x - 4 => x = 6 But the range was x ≥ 2 and 6 ≥ 2 => x = 6 is a SOLUTION

GMAT Problem Solving (PS) Questions

What is the sum of the different values of x such that |x+2| = 2|x-2|?

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