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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