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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`sqrt(128)`
To simplify the above square root, we must create a factor tree of the number inside the radical, defined as the radicand, and look for perfect squares that are apart of that factor tree. Since we are simplifying a square root, which means finding a number multiplied by itself to equal the number in the radicand, we are going to look for groups of 2 in the factorization. Therefore the number 128 can be factored down in this way:
128 = 64 x 2 = 8 x 8 x 2
To make this simpler, we can divide 128 by 2 to equal 64, which is a perfect square. 64 can then be broken down to equal a group of 8. Therefore factors that multiply together to equal 128 are 8 x 8 x 2.
Since there is a group of 8 with a single number (2) left over we can write the most simplified version of this radical in this form:
`8sqrt(2)`
The group of numbers always come outside the radical because you can evaluate the square root of that number (64 in this example). The single number (2) which is left over stays inside the radical because you cannot evaluate the square root of two.
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