GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Sep 2018, 22:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If |x + 1| = 2|x - 1|, what is the sum of the roots?

Author Message
TAGS:

### Hide Tags

Intern
Joined: 28 Aug 2016
Posts: 17
GMAT 1: 560 Q44 V23
If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

### Show Tags

Updated on: 17 Jul 2017, 08:08
2
6
00:00

Difficulty:

55% (hard)

Question Stats:

69% (01:28) correct 31% (01:48) wrong based on 260 sessions

### HideShow timer Statistics

If |x + 1| = 2|x - 1|, what is the sum of the roots?

A. 4
B. 6
C. 8
D. 20/3
E. 10/3

Originally posted by Amby02 on 17 Jul 2017, 03:03.
Last edited by chetan2u on 17 Jul 2017, 08:08, edited 2 times in total.
updated the OA
Math Expert
Joined: 02 Sep 2009
Posts: 49258
If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

### Show Tags

17 Jul 2017, 03:27
1
5
Amby02 wrote:
If |x + 1| = 2|x - 1|, what is the sum of the roots?

A. 4
B. 6
C. 8
D. 20/3
E. 10/3

Square the expression: $$x^2 + 2 x + 1=4 x^2 - 8 x + 4$$;

Re-arrange: $$3x^2-10x+3=0$$.

Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$:

$$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$.

Thus, for $$3x^2-10x+3=0$$, the sum of the roots is $$x_1+x_2=\frac{-b}{a}=\frac{-(-10)}{3}=\frac{10}{3}$$.

_________________
##### General Discussion
BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 3131
Location: India
GPA: 3.12
Re: If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

### Show Tags

17 Jul 2017, 07:31
1
There are 4 ways of representing the equation |x+2| = 2|x-1|

Case 1: |A| = |B| => A = B
(x+1) = 2(x-1)
x+1 = 2x - 2
Root 1 : x = 3

Case 2: |A| = |B| => A = -B
(x+1) = -2(x-1)
x+1 = -2x +2
3x = 1
Root 2 : x = $$\frac{1}{3}$$

Case 3: |A| = |B| => -A = B => A = -B(same as Case 2)
Case 4: |A| = |B| => -A = -B => A = B(sane as Case 1)

Since we have been asked the sum of the roots, it must be Root1(3) + Root2($$\frac{1}{3}$$)

The sum of the roots is $$\frac{10}{3}$$(Option E)
_________________

You've got what it takes, but it will take everything you've got

Math Expert
Joined: 02 Aug 2009
Posts: 6789
If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

### Show Tags

17 Jul 2017, 08:12
3
2
Amby02 wrote:
If |x + 1| = 2|x - 1|, what is the sum of the roots?

A. 4
B. 6
C. 8
D. 20/3
E. 10/3

Square both sides as both sides are positive..
$$x^2+2x+1=4x^2-8x+4..........3x^2-10x+3=0$$

no need to find the roots ..
SUM of the roots is $$\frac{-b}{a}=\frac{-(-10)}{3}=\frac{10}{3}$$
E
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Senior SC Moderator
Joined: 22 May 2016
Posts: 1977
If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

### Show Tags

17 Jul 2017, 09:02
1
Amby02 wrote:
If |x + 1| = 2|x - 1|, what is the sum of the roots?

A. 4
B. 6
C. 8
D. 20/3
E. 10/3

The only way the two sides will be equal is if quantities inside the absolute value brackets are 1) equal or 2) equal but with opposite signs. |x| = y or -y*

1. Remove the brackets

2. LHS = RHS or LHS = -RHS*

3. Set up the two equations

CASE 1: x + 1 = 2(x - 1) OR

CASE 2: x + 1 = -[2(x -1)]

4. Solve

CASE 1:
x + 1 = 2x - 2
3 = x ... x = 3

CASE 2:
x + 1 = -[2(x -1)]
x + 1 = -(2x - 2)
x + 1 = -2x + 2
3x = 1
x = $$\frac{1}{3}$$

5. Check x = 3 and x = $$\frac{1}{3}$$ When removing absolute value brackets, I always check to see whether or not the roots satisfy the original equation. Both work.

6. Sum of roots is 3 + $$\frac{1}{3}$$ = $$\frac{10}{3}$$

*(and |y| = x or -x). You can reverse RHS and LHS, where RHS = LHS or RHS = - LHS. pushpitkc shows the four possibilities. The latter two are identical to the first two.
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

Director
Joined: 21 Mar 2016
Posts: 532
Re: If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

### Show Tags

18 Jul 2017, 07:55
Amby02 wrote:
If |x + 1| = 2|x - 1|, what is the sum of the roots?

A. 4
B. 6
C. 8
D. 20/3
E. 10/3

whenever there is absolute sign on both sides,,square it off...
gives (x+1)^2 = 4 (x-1)^2
x^2 + 2x + 1 = 4x^2 + 4 - 8x
3x^2 - 10x + 3 = 0
sum of both the roots = -b/a = 10/3
Intern
Joined: 14 Sep 2017
Posts: 4
Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?  [#permalink]

### Show Tags

Updated on: 20 Jan 2018, 09:55
1
Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?

A) 4
B) 6
C) 8
D) 20/3
E) 10/3

Originally posted by fraserlawson on 20 Jan 2018, 09:48.
Last edited by Bunuel on 20 Jan 2018, 09:55, edited 1 time in total.
Renamed the topic, edited the question
CEO
Joined: 12 Sep 2015
Posts: 2856
Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?  [#permalink]

### Show Tags

Updated on: 20 Jan 2018, 11:40
1
Top Contributor
fraserlawson wrote:
Which of these is the sum of the solutions of |x+1|=2|x-1|?

A) 4
B) 6
C) 8
D) 20/3
E) 10/3

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

Given: |x+1|=2|x-1|

So, EITHER x+1=2(x-1) OR x+1=-[2(x-1)]

x+1=2(x-1)
Expand: x + 1 = 2x - 2
Solve: x = 3

x+1=-[2(x-1)]
Expand: x + 1 = -2x + 2
Solve: x = 1/3

When we plug the solutions into the original equation, we find that there are no extraneous roots

So, the SUM of the solutions = 3 + 1/3
= 3 1/3
= 10/3

RELATED VIDEO

_________________

Brent Hanneson – GMATPrepNow.com

Originally posted by GMATPrepNow on 20 Jan 2018, 09:58.
Last edited by GMATPrepNow on 20 Jan 2018, 11:40, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 49258
If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

### Show Tags

20 Jan 2018, 10:00
1
1
Intern
Joined: 14 Sep 2017
Posts: 4
Re: Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?  [#permalink]

### Show Tags

20 Jan 2018, 11:03
GMATPrepNow wrote:

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

Given: |x+1|=2|x-1|

So, EITHER x+1=2(x-1) OR EITHER x+1=-[2(x-1)]

x+1=2(x-1)
Expand: x + 1 = 2x - 2
Solve: x = 3

x+1=-[2(x-1)]
Expand: x + 1 = -2x + 2
Solve: x = 1/3

When we plug the solutions into the original equation, we find that there are no extraneous roots

So, the SUM of the solutions = 3 + 1/3
= 3 1/3
= 10/3

Awesome way to solve the problem, and way faster than what I did to brute-force it. Thank you!

One question, though. How is this approach different from how one would solve |x+1|=2(x-1)? Or x+1 = 2|x-1|?

I ask because 2|x-1|-(x+1)=0 has roots of 1/3 and 3, whereas 2(x-1)-|x+1|=0 only has the root at 3 (unless I've used Wolfram wrong). Is there something about knowing which one of the two absolute value elements to pick?
CEO
Joined: 12 Sep 2015
Posts: 2856
Re: Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?  [#permalink]

### Show Tags

20 Jan 2018, 11:40
Top Contributor
fraserlawson wrote:
GMATPrepNow wrote:

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

Given: |x+1|=2|x-1|

So, EITHER x+1=2(x-1) OR x+1=-[2(x-1)]

x+1=2(x-1)
Expand: x + 1 = 2x - 2
Solve: x = 3

x+1=-[2(x-1)]
Expand: x + 1 = -2x + 2
Solve: x = 1/3

When we plug the solutions into the original equation, we find that there are no extraneous roots

So, the SUM of the solutions = 3 + 1/3
= 3 1/3
= 10/3

Awesome way to solve the problem, and way faster than what I did to brute-force it. Thank you!

One question, though. How is this approach different from how one would solve |x+1|=2(x-1)? Or x+1 = 2|x-1|?

I ask because 2|x-1|-(x+1)=0 has roots of 1/3 and 3, whereas 2(x-1)-|x+1|=0 only has the root at 3 (unless I've used Wolfram wrong). Is there something about knowing which one of the two absolute value elements to pick?

Good question.
When there are absolute values on BOTH sides, it doesn't matter which one you choose to make negative.

For example, with |x+1|=2|x-1|, we could have also gone with:
EITHER x+1=2(x-1) OR -(x+1)=2(x-1)
The second equation still results in x = 1/3

Cheers,
Brent
_________________

Brent Hanneson – GMATPrepNow.com

Re: Which of these is the sum of the solutions of |x + 1| = 2|x - 1|? &nbs [#permalink] 20 Jan 2018, 11:40
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.