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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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73% (02:06) correct
27%
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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 [ 15.36 KiB | Viewed 27313 times ]
A. \(4\)
B. \(\frac{16}{3}\)
C. \(6\)
D. \(\frac{20}{3}\)
E. \(8\)
Attachment:
2023-12-04_21-00-05.png [ 15.36 KiB | Viewed 27313 times ]
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Adrian_D
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.
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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.
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.
General Discussion
gmatophobia
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hugogva
\(|x+2| = 2|x-2|\)
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
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
rakman123
---------
^ 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?
Bunuel
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?
Why do you not test the second equation with a negative? Shouldn't this result in 4 scenarios?
pisker
Your doubt is addressed here: https://gmatclub.com/forum/what-is-the- ... l#p3310573
Hope this helps.
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Bunuel
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)
Nevermind - I see it now thanks!
Hey, can you explain. Didn't quite understadn why didn't we add up all the 4 values
pisker
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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
|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
Gaurav07
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
=> Sum of different values of x = 2/3 + 6 = 20/3
So, Answer will be D
Hope it helps!
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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!
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hugogva
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"
Attachment:
Screenshot 2024-10-19 at 7.54.03 PM.png [ 36.58 KiB | Viewed 18822 times ]
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
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Tip : Whenever you see |a|=|b| type Question : SOLVE it by Squaring both sides
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Bunuel, we are testing only for same signs(+ or -) and opposite sides BECAUSE it's a pair? as in |x+2| and |x-2|? if the absolute value wasn't like this, we should test for 4 possible cases as we do?
Bunuel
SwethaReddyL
The reason is not that these two expressions form a special “pair.” The reason is that the equation has the form \(|A| = |B|\). In such cases, only two relative-sign cases are needed: \(A = B\) or \(A = -B\). We do not need four separate cases because the same-sign cases are equivalent to each other, and the opposite-sign cases are also equivalent to each other.
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Hi SwethaReddyL,
Yes, your instinct is exactly right, and let me make it precise.
You're reacting to Venu01298's point that the "both positive" and "both negative" versions give the same solutions, as do the two "opposite sign" versions. Here's why that happens.
Why 4 collapses to 2
Your equation has an absolute value on both sides - it's the form |A| = |B|, where here A = x+2 and B = 2(x-2). Lay out all four sign choices:
- (+, +): A = B
- (-, -): -A = -B -> multiply both sides by -1 -> A = B (same as the first)
- (+, -): A = -B
- (-, +): -A = B -> multiply both sides by -1 -> A = -B (same as the third)
So the four cases fold into just two: A = B or A = -B. The reason is exactly the "pair" you spotted - because both sides are wrapped in absolute values, flipping both signs is the same equation. That's the general rule:
|A| = |B| is true exactly when A = B or A = -B.
Your second question
If the structure were different, the collapse can fail. The clean trick only works when the whole equation is one absolute value equal to another. If you had something like |x+2| = 2|x-2| + 5 (an extra loose term), negating the whole equation no longer reproduces it, so the cases stay separate - there you'd fall back to the number-line region method (the one BrushMyQuant used: split at the critical points and test each interval).
Lock it in
Try the simpler |x| = |x - 4|:
- x = x - 4 -> no solution
- x = -(x - 4) -> 2x = 4 -> x = 2
Write out the other two sign versions and you'll see they reproduce these same two - never anything new. The pairing always does the collapsing for you.
Answer: D
_________________
Yes, your instinct is exactly right, and let me make it precise.
You're reacting to Venu01298's point that the "both positive" and "both negative" versions give the same solutions, as do the two "opposite sign" versions. Here's why that happens.
Why 4 collapses to 2
Your equation has an absolute value on both sides - it's the form |A| = |B|, where here A = x+2 and B = 2(x-2). Lay out all four sign choices:
- (+, +): A = B
- (-, -): -A = -B -> multiply both sides by -1 -> A = B (same as the first)
- (+, -): A = -B
- (-, +): -A = B -> multiply both sides by -1 -> A = -B (same as the third)
So the four cases fold into just two: A = B or A = -B. The reason is exactly the "pair" you spotted - because both sides are wrapped in absolute values, flipping both signs is the same equation. That's the general rule:
|A| = |B| is true exactly when A = B or A = -B.
Your second question
If the structure were different, the collapse can fail. The clean trick only works when the whole equation is one absolute value equal to another. If you had something like |x+2| = 2|x-2| + 5 (an extra loose term), negating the whole equation no longer reproduces it, so the cases stay separate - there you'd fall back to the number-line region method (the one BrushMyQuant used: split at the critical points and test each interval).
Lock it in
Try the simpler |x| = |x - 4|:
- x = x - 4 -> no solution
- x = -(x - 4) -> 2x = 4 -> x = 2
Write out the other two sign versions and you'll see they reproduce these same two - never anything new. The pairing always does the collapsing for you.
Answer: D
SwethaReddyL
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|x + 2| = |2x -4|
Squaring both sides and using a^2-b^2 formula
(x+2+2x-4)(x+2-2x+4) = 0
(3x-2)(6-x) = 0
x = 2/3 and x = 6
adding both we get 20/3
Squaring both sides and using a^2-b^2 formula
(x+2+2x-4)(x+2-2x+4) = 0
(3x-2)(6-x) = 0
x = 2/3 and x = 6
adding both we get 20/3
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