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

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

RELATED VIDEO

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

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.

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

10. Absolute Value

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Hope it helps.

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

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

Why would the first region not be valid?

BrentGMATPrepNow its me again

I have a question about highlighted part in the post above
so basically when i need to check the region -1 < x < 1, (when x is in between) it doesnt matter which of two sides i make negative
|x + 1| = 2|x - 1| Right ? -(x + 1)= 2(x - 1) or (x + 1)= -2(x - 1) or maybe it is a case with this problem ?
the point is that i considered all 4 cases and got 3+3+1/3+1/3

What if the question asked "how many solutions does the equation |x + 1| = 2|x - 1| have ?

thank you

chetan2u how do you know that x must be positive so as you can square both sides or may be you mean value inside absolute value brackets ? so i can square absolute value inside absolute value brackets anytime i see it right ?

Hi
Let me give you an explanation which will help in solving most of the Modulus problems since squaring only works when both the sides contain modulus.
Mod is nothing but distance from number line
|X+1| = |X- (-1)|
|X-1 is fine|= |X-1|
Always write the expression under the mod like above ( Something like x minus constant)
Now mark -1 and 1 on the number line : There will be three regions
1st region = Less than -1 = Take any integer which makes the calculation easier say -2
Then equation will be -(x+1) = -2(X-1)
Solving , x = 3
2nd Region = Between -1 and 1 = Take any integer which makes the calculation easier say 0
Then equation will be (x+1)=-2(x-1)
Solving , x= 1/3
3rd region = Beyond 1 = Take any integer which makes the calculation easier say 2
Then equation will be (x+1) = 2(x-1)
Solving , x= 3 (same as region 1)
So sum of roots = 1/3 + 3 = 10/3

NOTE : This method works even when there are 3 modulus equations or inequalities or anything - It is just a fantastic concept I wish to share

ibra10 thank you

using your concept how would you solve this one |X+4| > |X- 2|

dave13

Hi Dave
Step 1 : Write the inequalities as follows so as to plot on number line |x-(-4)| > |x-2|
We will again have 3 regions : 1st Region = <2 | 2nd Region = Between 2 and 4 | 3rd Region = Greater than 4

Solving for region 1 = Region less than 2 = Take x=1
For Region 1 , the inequality will be : (x+4) > -(x-2)
Solving the above gives us : X > -1

Solving for Region 2 = Region between 2 and 4 = Take x=3
For Region 2 , the inequality will be (x+4) > (x-2)
X will cancel out thus , no values for region 2

Similarly for Region 3 = Region beyond 4 = Take x=5
For Region 3 , the inequality will again be (x+4)>(x-2)
X will cancel out again , no values for region 3

So answer is X>-1

Note : You can try to put any value for X greater than -1 and the inequality will be satisfied

Always happy to help

Hi

The mod are always positive, so we can square both sides.
In my signature you will find ways to tackle these questions through 3 methods including the critical method as shown above.

Hi chetan2u
I solved this question by critical value method and got the answer ,and I'm doing great with this approach.
However , I am confused with the squaring method as in when and when not to use it because on some questions this approach gives the right answer very quickly , but on others it makes the equation more complex.
Can you please tell on what type of mod questions the squaring method is easy to apply?

Wherever you are sure that the both the sides of equation are positive, you can square it.