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30 Aug 2006, 04:10
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Is (x-4)^2>9-(x-3)^2?
(1) When x-2 is divided by 9, the remainder is 8.
(2) |x-8|>|x-3|.
(1) When x-2 is divided by 9, the remainder is 8.
(2) |x-8|>|x-3|.
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30 Aug 2006, 08:27
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So you can reduce the question stem to (x-8)(x+1)>0
So the question now reads Is (x-8)(x+1)>0? Get the critical pts
x<-1 -1<x<8 x>8 of these only x<-1 and x>8 satisfy the inequality.
So the question now reads is x<-1 or x>8
From 1 x-2=9a+8 --> x=9a+10 or x=9a+1 So x could be 1 or 10 ,19 etc insufficient
From 2 |x-8|>|x-3|.the critical pts are x<3 3<x<8 x>8 here both x<3 and 3<x<8 satisfies the inequality again insufficient.
Combining both only x=1 works unless the remainder of a negative number -8 when divided by 9 is considered 8 not -8 So I think its C
So the question now reads Is (x-8)(x+1)>0? Get the critical pts
x<-1 -1<x<8 x>8 of these only x<-1 and x>8 satisfy the inequality.
So the question now reads is x<-1 or x>8
From 1 x-2=9a+8 --> x=9a+10 or x=9a+1 So x could be 1 or 10 ,19 etc insufficient
From 2 |x-8|>|x-3|.the critical pts are x<3 3<x<8 x>8 here both x<3 and 3<x<8 satisfies the inequality again insufficient.
Combining both only x=1 works unless the remainder of a negative number -8 when divided by 9 is considered 8 not -8 So I think its C
