N^2 + 1

Question

Hi,

This is from Veritas Advanced Word Problems and Quantitative Review.

Which of the following describes all values for which n^2-1 >= 0?

A. N>=1
B. N<=-1
C. 0<=n<=1
D. N<=-1 or N>=1
E. -1<=n<=1

I got the answer right by deduction, but the straight algebra doesn't seem to work for me. Can anyone explain what I am doing wrong?

My algebra went like this:
n^2 - 1 >=0
(n-1)(n+1) >=0
n-1 >=0 or n+1 >=0
n>=1 or n>=-1

However, the second condition is clearly not right (n cannot be 0 for instance). I am not multiplying or dividing this inequality by a negative a number, so there should be no reason to switch the sign for the inequality. What is the proper way to do the algebra?

Answer is D.

limestone · Answer

n^2 -1 >=0 means
n^2 >=1 or
Abs(n) >=1, then n<=-1 or n>=1
Pick D

My algebra went like this:
n^2 - 1 >=0
(n-1)(n+1) >=0
n-1 >=0 or n+1 >=0
n>=1 or n>=-1

There're two cases here:
1. n-1>=0 and n+1>=0(not or because both n+1 and n-1 need to be positive to have their product a positive, if n+1 is positive and n-1 is negative (in case n = 0.5), then the condition will not be met)
=> n>=1 and n>=-1 => n>=1 ( as when n>=1, n is obviously larger than -1)
2. n-1<=0 and n+1<=0 (and because of the same reason in case 1)
=> n<=1 and n<=-1 => n<=-1 (as when n<=-1, n is obviously smaller than 1)
Combine the two cases n>=1 or n<=-1

Jim@Grockit · Answer

skalevar wrote:Hi,

This is from Veritas Advanced Word Problems and Quantitative Review.

Which of the following describes all values for which n^2-1 >= 0?

A. N>=1
B. N<=-1
C. 0<=n<=1
D. N<=-1 or N>=1
E. -1<=n<=1

I got the answer right by deduction, but the straight algebra doesn't seem to work for me. Can anyone explain what I am doing wrong?

My algebra went like this:
n^2 - 1 >=0
(n-1)(n+1) >=0
n-1 >=0 or n+1 >=0
n>=1 or n>=-1

However, the second condition is clearly not right (n cannot be 0 for instance). I am not multiplying or dividing this inequality by a negative a number, so there should be no reason to switch the sign for the inequality. What is the proper way to do the algebra?

Answer is D.

The issues are that individually, (n-1) and (n+1) can be both negative (say n= -3) or both positive and still have their product >=0. Because they can both be negative, you can't logically go to your 3rd step "or".

Also what limestone said.

Rahul@gurome · Answer

Your solution till second step is correct which is (n-1)(n+1)>= 0.

This means case (1) both (n-1) >= 0 and (n+1)>= 0.
Or case (2) both (n-1) < = 0 and (n+1) < = 0.

Take case (1).
It means n >= 1 and n >= -1.
Note the word and here. Both the conditions n >= 1 and n >= -1 have to be satisfied. This is only possible if n > = 1. For example you cannot take n = -1/2. Though -1/2 >= -1, it is not >= 1.
What you did was n >= 1 or n >= -1 which is not correct.
Hence case (1) gives us that n>=1.

Take case (2).
It means n <= 1 and n < = -1.
Do the same thing. On combining we have that n <= -1. For example you cannot take n = ½.
Though ½ <= 1, it is not <= -1.
Hence case (2) gives us that n <= -1.

So we can say that n^2 – 1 >= 0 implies n >= 1 or n <= -1.

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

skalevar · Answer

Thanks all. that makes a lot more sense!

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