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.
_________________

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.
_________________

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
_________________

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
_________________

x+1=2x-2
x=3

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

Answer E. 10/3

THIS PROBLEM IS EASY

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

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


_________________

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 ?

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
_________________

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!

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
_________________

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.

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

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??

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

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

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.
_________________

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

Thank you very much