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

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

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New post 10 Jul 2018, 04:15
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Which of the following is the sum of solutions of |x+1| = 2|x-1|?

A. 4
B. 6
C. 8
D. \(\frac{20}{3}\)
E. \(\frac{10}{3}\)
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Re: Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 10 Jul 2018, 04:33
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2
Karthik200 wrote:
Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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


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

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New post 11 Jul 2018, 22:36
Karthik200 wrote:
Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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


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.
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Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 16 Jul 2018, 09:51
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Scenerio 1: x + 1 = 2(x - 1) Implies that x = 3.
Scenerio 2: (x + 1) = -2(x - 1) Implies that x = 1/3. [The reason we cannot use a negative sign on both the sides when we are equating in scenerio 2 is then there will not be any difference between scenario 1 and 2]

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

Option E is the correct answer
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Re: Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 16 Jul 2018, 10:26
Karthik200 wrote:
Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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



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

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New post 18 Jul 2018, 20:03
x+1=2x-2
x=3

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

Answer E. 10/3

THIS PROBLEM IS EASY
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Re: Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 17 Dec 2018, 12:49
Bunuel wrote:
Karthik200 wrote:
Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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


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.



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

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New post 18 Dec 2018, 02:07
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Hi dave13,

You are absolutely correct here -
Quote:
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.


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.
Quote:
|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


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 wrote:
Bunuel wrote:
Karthik200 wrote:
Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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


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.



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

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

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New post 18 Dec 2018, 03:06
1
Gladiator59 wrote:
Hi dave13,

You are absolutely correct here -
Quote:
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.


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.
Quote:
|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


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




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

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New post 18 Dec 2018, 05:43
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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
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Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 21 Mar 2019, 20:18
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!
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Re: Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 22 Mar 2019, 00:06
1
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. :-)

jfranciscocuencag wrote:
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!


Posted from my mobile device
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Re: Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 16 Apr 2019, 16:45
Bunuel wrote:
Karthik200 wrote:
Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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


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.


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

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New post 17 Apr 2019, 03:33
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
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Re: Which of the following is the sum of solutions of |x+1| = 2|x-1|?  [#permalink]

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New post 28 Apr 2019, 01:19
Bunuel wrote:
Karthik200 wrote:
Which of the following is the sum of solutions of |x+1| = 2|x-1|?

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


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.


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

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New post 29 Apr 2019, 09:23
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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
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Re: Which of the following is the sum of solutions of |x+1| = 2|x-1|?   [#permalink] 29 Apr 2019, 09:23
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