Please solve `sqrt 128` , and show your work.

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`sqrt128`

First, let's express 128 with its prime factors.

`=sqrt(2^7)`

Using the property of exponent which is `a^m*a^n = a^(m+n)` , we may re-write `2^7` as:

`=sqrt(2^2*2^2*2^2*2)`

Applying the product property of square root which is `sqrt(a*c) = sqrta * sqrtc` , we then have:

`= sqrt(2^2) *sqrt(2^2)*sqrt(2^2)*sqrt2`

Note that `sqrt(a^2) = a` .  So.

`=2*2*2*sqrt2`

`=8sqrt2`

Hence, `sqrt128=8sqrt2` .

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