Which of the following is the sum of solutions of |x+1| = 2|x-1|?

Question

A. 4
B. 6
C. 8
D.  \frac{20}{3} 
E.  \frac{10}{3}

Bunuel · Accepted Answer

Square the given equation:

(x + 1)^2 = 4(x - 1)^2 ;

4(x - 1)^2 - (x + 1)^2 = 0 ;

Apply a^2 - b^2 = (a - b)(a + b) to the above:  (2(x - 1) - (x + 1))(2(x - 1) + (x + 1)) = 0 ;

(x - 3)(3x - 1) = 0 ;

x = 3  or  x= \frac{1}{3} .

The sum of the roots  = 3 + \frac{1}{3} = \frac{10}{3} .

Answer: E.

OR: |x+1| = 2|x-1| can be expanded either the way that both LHS and RHS have the same sign or different signs. So

Either: x + 1 = 2(x - 1) --> x = 3.
Or: x + 1 = -2(x - 1) --> x = 1/3.

The sum of the roots  = 3 + \frac{1}{3} = \frac{10}{3} .

Answer: E.

CAMANISHPARMAR · Answer

Scenerio 1: x + 1 = 2(x - 1) Implies that x = 3.
Scenerio 2: (x + 1) = -2(x - 1) Implies that x = 1/3.

Sum of possible solutions is  \frac{10}{3}

Option E is the correct answer

lfc · Answer

Hi,

Squaring both sides will do the trick. 
I started plugging in values into  |x+1| / /|x-1|=2  and eliminating options , took 2+ mins.

GyMrAT · Answer

Given: |x+1| = 2|x-1| For x+1 > 0 & x-1 > 0, we have x>-1 & x>1, we get x+1 = 2(x-1), hence x = 3 > -1, is within range. For x+1 < 0 & x-1 < 0, we have x<-1 & x<1, we get -(x+1) = -2(x-1), hence x = 3 > 1, is not acceptable For x+1 > 0 & x-1 < 0, we have x>-1 & x<1, we get x+1 = -2(x-1), hence x = 1/3 < 1 & > -1, is within range. For x+1 < 0 & x-1 > 0, we have x<-1 & x>1, we get -(x+1) = 2(x-1), hence x = 1/3 < 1, is not acceptable. Hence, we have, x = 3 & 1/3 Sum of solutions = 3 + 1/3 = 10/3 Answer E. Thanks, GyM

gmatzpractice · Answer

x+1=2x-2
x=3

-x-1=2x-2
x=1/3

Answer E. 10/3

THIS PROBLEM IS EASY

dave13 · Answer

Gladiator59   Bunuel   pushpitkc  how do we know that x is not negative before we square..... they say that dont square unless you know that variable is not negative.

Also why my solution below didnt work ?

|x+1| = 2|x-1|

when x is positive

x+1=2x-2 
x =3

When x is negative

|x+1| = 2|x-1|

x-1=2x-2

x = 1

3+1 = 4 = sum of solutions

Gladiator59 · Answer

Hi  dave13 ,

You are absolutely correct here -

Sine both LHS and RHS are mod expressions they are always nonegative. ( definiton of mod) and hence we can square here.

Also, in your solution with x is negative part you have made calculation errors.

it should be as Bunuel points out.

x + 1 = - ( 2(x-1))
x + 1 = -2x + 2
3x = 1
x = 1/3

So it will be 3 + 1/3 or 10/3

Hope this is clear now.

Best,
Gladi

dave13 · Answer

many thanks Gladiator59

optionally I could mame LHS negative right ? -|x+1| for example like this -(x+1) = 2(x-1) -x-1=2x-2 1 = 3x 3x=1 x = 1/3 So my mistake is that I opened moduli incorrectly, by not putting minus sign in front of brackets, but instead I simply changed sign in the brackets ? right ?

Gladiator59 · Answer

dave13 , absolutely. You just have to go with both signs ( outside bracket)

So you calculated correctly for positive and for negative you erred with the calculation. Either a = -b or -a =b of these is the same and correct way.

Best,
Gladi

jfranciscocuencag · Answer

Gladiator59 
 dave13

Hello!

Could someone please explain to me why do we have to change the -ve just to one side?

|x+1| = 2|x-1|

+ve way... x + 1 = 2x - 2

And I thought the -ve way was:

-ve way...

(-)|x+1| = 2(-)|x-1|  =

- x- 1 = 2(-x+1)  =

- x- 1 = -2x + 2

Kind regards!

Gladiator59 · Answer

Just think about it. A negative sign on both sides will cancel out and the second case will be identical to the first one. Hence the assymetrical negative sign to take care of the case where one mod evaluates as negative and the other as positive. There will be four cases actually. But two of those cases will be identical to the other two. +M1 +M2 -M1 -M2 And +M1 -M2 -M1 +M2 Notice how the sets are identical and considering only one from each is sufficient. Hope this helps.

Posted from my mobile device

edlc313 · Answer

Bunuel

Can you explain why you squared both sides to solve this problem? I cannot figure out why you figured this was the quickest approach.

mariossid · Answer

The solution is:
(x+1)^2 = 4(x-10)^2 <=> 3x^2 -10x -5=0 
From Vieta’s formula we know that x1 + x2 = -b/a which is 10/3
E

venkivety · Answer

learned a new technique when the mod is given..
squaring on the both sides,,
thanx bunuel..

bt if the mod is on only one side of the equation???
and can the squaring be done when inequalities vth mod??

saukrit · Answer

Another simple solution may be:

Squaring both sides we get
 
 (x+1)^2= 4(x-1)^2

3x^2-10*x+3=0 
Sum of roots of a quadratic equation = -b/a

Hence answer = 10/3

Correct answer, E

nisthagupta28 · Answer

I solved it by using critical value. Got the correct answers. 
Just wanted to know if that is one of approaches to follow?

Posted from my mobile device

CEdward · Answer

Learned something with squaring both sides. Is that something we can do for all questions of this flavour?

I solved this way:

Take the positive
| x + 1 | = 2 | x - 1 | 
x + 1 = 2x - 2
x = 3

Take the negative
| x + 1 | = 2 (-) | x - 1 | 
x + 1 = -2x + 2
x = 1/3

3 + 1/3 = 10/3

Answer is E.

jcerdae · Answer

Can we do the square on both sides if there is MOD in only 1 side?

Thank you very much

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Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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