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Range

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Intern
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Joined: 25 Oct 2010
Posts: 43

Kudos [?]: 154 [0], given: 13

WE 1: 3 yrs
Range [#permalink]

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New post 11 Nov 2010, 07:01
I have a doubt regarding range

If y^2 <64, then what is the range in which y exists

Answer: -8<y<8

I want to know the steps involved in solving this. And the logic.
Thanks or the help.

Kudos [?]: 154 [0], given: 13

Manager
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Joined: 30 Sep 2010
Posts: 56

Kudos [?]: 60 [0], given: 0

Re: Range [#permalink]

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New post 11 Nov 2010, 09:24
y^2 < 64
OR |y| < 8
OR -8< y< 8

Kudos [?]: 60 [0], given: 0

Manager
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Joined: 01 Nov 2010
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Location: Zürich, Switzerland
Re: Range [#permalink]

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New post 11 Nov 2010, 12:35
1. Square of a negative or positive number is always positive.
2. Range of a set of numbers is defined as the greatest value in the data set minus the least value.
This indicates, y must fall within the +ve and -ve limits of 8.

Thus, the range of y should be 7 -(-7) = 14.

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Re: Range [#permalink]

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New post 11 Nov 2010, 14:18
student26 wrote:
I have a doubt regarding range

If y^2 <64, then what is the range in which y exists

Answer: -8<y<8

I want to know the steps involved in solving this. And the logic.
Thanks or the help.


\(y^2<64\)
\(y^2-64<0\)
\((y+8)(y-8)<0\)

Such an expression will be less than 0, if one term is negative and the other is positive. This will only happen when y is between 8 and -8 (Below -8, both terms are negative and above 8, both terms are positive)

Hence the range of \(y\) is \(-8<y<8\)
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Kudos [?]: 1186 [0], given: 25

Re: Range   [#permalink] 11 Nov 2010, 14:18
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