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

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

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New post Updated on: 17 Jul 2017, 08:08
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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
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If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

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New post 17 Jul 2017, 03:27
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6
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 \(x_1+x_2=\frac{-b}{a}=\frac{-(-10)}{3}=\frac{10}{3}\).

Answer: E.
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Re: If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

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New post 17 Jul 2017, 07:31
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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)
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If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

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New post 17 Jul 2017, 08:12
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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
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If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

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New post 17 Jul 2017, 09:02
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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 + \(\frac{1}{3}\) = \(\frac{10}{3}\)

Answer D

*(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.
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Re: If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

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New post 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
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Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?  [#permalink]

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

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New post Updated on: 20 Jan 2018, 11:40
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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

Answer: E

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

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New post 20 Jan 2018, 10:00
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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


10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Re: Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?  [#permalink]

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New post 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

Answer: E


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?
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Re: Which of these is the sum of the solutions of |x + 1| = 2|x - 1|?  [#permalink]

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New post 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

Answer: E


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
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Re: If |x + 1| = 2|x - 1|, what is the sum of the roots?  [#permalink]

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New post 23 Jan 2019, 11:15
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


key points at which this inequality will lie will be

x = -1 or x =1

This will divide 3 regions on the number line,
------(-1)------(1)--------

now when x < -1, This region is not valid

-x-1 = -2x + 2
x = 3, Not applicable

-1 < x < 1, this region is valid

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

x > 1, this region is valid

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

(a) + (b)
10/3

Answer E
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Re: If |x + 1| = 2|x - 1|, what is the sum of the roots?   [#permalink] 23 Jan 2019, 11:15
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