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Bunuel
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Kinshook
Bunuel
What is 4x^2 ?

(1) |2x – 2| + |2x + 2| = 4
(2) |3x – 1| + |3x + 1| = 6


Q: What is 4x^2 ?

#1:
|2x – 2| + |2x + 2| = 4
|x-1|+|x+1| =2
-1<=x<=1
NOT SUFFICIENT

#2:
|3x – 1| + |3x + 1| = 6
\(|x-\frac{1}{3}| + |x+\frac{1}{3}| = 2\)
x=1 or x = -1
=> x^2 = 1
=> 4x^2 = 4
SUFFICIENT

IMO B

Can you elaborate statement 1? Also, Why are we getting a range in #1 but not in #2?
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Hello!

Can you please explain statement 2?


Kinshook
Bunuel
What is 4x^2 ?

(1) |2x – 2| + |2x + 2| = 4
(2) |3x – 1| + |3x + 1| = 6


Q: What is 4x^2 ?

#1:
|2x – 2| + |2x + 2| = 4
|x-1|+|x+1| =2
-1<=x<=1
NOT SUFFICIENT

#2:
|3x – 1| + |3x + 1| = 6
\(|x-\frac{1}{3}| + |x+\frac{1}{3}| = 2\)
x=1 or x = -1
=> x^2 = 1
=> 4x^2 = 4
SUFFICIENT

IMO B
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AmanMatta
Kinshook
Bunuel
What is 4x^2 ?

(1) |2x – 2| + |2x + 2| = 4
(2) |3x – 1| + |3x + 1| = 6


Q: What is 4x^2 ?

#1:
|2x – 2| + |2x + 2| = 4
|x-1|+|x+1| =2
-1<=x<=1
NOT SUFFICIENT

#2:
|3x – 1| + |3x + 1| = 6
\(|x-\frac{1}{3}| + |x+\frac{1}{3}| = 2\)
x=1 or x = -1
=> x^2 = 1
=> 4x^2 = 4
SUFFICIENT

IMO B

Can you elaborate statement 1? Also, Why are we getting a range in #1 but not in #2?

Please solve the equations by regions or distances.
For test, put x=0 in #1 and x=0 in #2 to confirm.

Posted from my mobile device
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kunalbean,
solve statement 1: |x+1| + |x-1|=2
x>1 yields x=1
x<-1 x=-1
for the range -1<=x<=1 you get 2=2 meaning that for this range any value of x satisfies the equation (x is not bound to be an integer).
hope it helped
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Hello afa13

Thank you for the response. The range is defined for -1<=x=>1. When squared -1=1=1. Hence it should be sufficient ?
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AmanMatta
Kinshook
Bunuel
What is 4x^2 ?

(1) |2x – 2| + |2x + 2| = 4
(2) |3x – 1| + |3x + 1| = 6


Q: What is 4x^2 ?

#1:
|2x – 2| + |2x + 2| = 4
|x-1|+|x+1| =2
-1<=x<=1
NOT SUFFICIENT

#2:
|3x – 1| + |3x + 1| = 6
\(|x-\frac{1}{3}| + |x+\frac{1}{3}| = 2\)
x=1 or x = -1
=> x^2 = 1
=> 4x^2 = 4
SUFFICIENT

IMO B

Can you elaborate statement 1? Also, Why are we getting a range in #1 but not in #2?

AmanMatta

For statement 1, you can put x=0, x=1/2, x=-1/2, then you find that range is correct.

For me, drawing the graph of |x-1| and |x+1| makes easier to understand such range solution.

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