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Source: — Data Sufficiency |
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by gmatmachoman » Mon Feb 15, 2010 10:49 pm
IMO C
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by shashank.ism » Mon Feb 15, 2010 11:28 pm
austin wrote:If f(x) = x^n - a = 0, how many REAL roots does x have?

(1) a < 0
(2) n = 2k, where k is a positive integer
x^n - a = 0 --> x= a^(1/n)
so there can be many solution depending on a and n
St. 1: a<0 i.e. -ve values then also many soln. --------------not sufficient
St. 2: n=2k then also many soln. --------------not sufficient
combined : since a is -ve and and we are taking nth root where n is even (as n=2k)
so there will be no solution
Hence, it is sufficient Ans C
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by ajith » Mon Feb 15, 2010 11:59 pm
austin wrote:If f(x) = x^n - a = 0, how many REAL roots does x have?

(1) a < 0
(2) n = 2k, where k is a positive integer
1) a<0
is not sufficient
Say when n = 1 and a= -3
x+3 has one root x = -3
when n =2 and a= -3
x^2+3 = 0 doesn't have real roots

2) n = 2k is not sufficient

say k = 1 and a = 1
it has 2 roots
x^2-1 =0 has 2 roots

say k = 1 and a = -1
x^2+1=0 has 0 roots

Combining

x^2k + (a positive number) =0 doesnt have any real roots
So, C
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by Ian Stewart » Tue Feb 16, 2010 12:55 am
austin wrote:If f(x) = x^n - a = 0, how many REAL roots does x have?

(1) a < 0
(2) n = 2k, where k is a positive integer
There's an issue with the wording here. When the question uses the word 'roots', what it means is 'solutions to the equation'. It makes no sense to talk about how many 'roots' x has in that case; they mean to ask how many roots 'x^n - a = 0' has, or as is sometimes said, how many roots f(x) has. If you simply talk about the 'roots' of a number or letter, it would be natural to take that to mean square roots, cube roots, and so on, which is not at all what the question is talking about. If the wording in the post above is that used in the source, then there are problems with that source.
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by austin » Tue Feb 16, 2010 3:12 am
Ian,

You are right, it has to be f(x).
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by shashank.ism » Sun Feb 21, 2010 9:03 pm
Ian Stewart wrote:
austin wrote:If f(x) = x^n - a = 0, how many REAL roots does x have?

(1) a < 0
(2) n = 2k, where k is a positive integer
There's an issue with the wording here. When the question uses the word 'roots', what it means is 'solutions to the equation'. It makes no sense to talk about how many 'roots' x has in that case; they mean to ask how many roots 'x^n - a = 0' has, or as is sometimes said, how many roots f(x) has. If you simply talk about the 'roots' of a number or letter, it would be natural to take that to mean square roots, cube roots, and so on, which is not at all what the question is talking about. If the wording in the post above is that used in the source, then there are problems with that source.
ian that was a nice explanation related to roots..will keep it in mind.
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by girish3131 » Tue Mar 09, 2010 1:21 am
IMO A

b'cos ques ans how many real roots r thr..

so option A says that NO SOLUTION.... so that mean we get ans A

ta
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