How do you simplify square root of 175 minus square root of 28?

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Answer 1 · 2015-07-29T17:01:21

#sqrt175-sqrt28 = 3sqrt7#

First, simplify each square root:

#sqrt175#, is #175# divisible by a perfect square (other than 1)?

We could go through the list starting with #2^2 = 4# is not a factor and #3^2 = 9# is not a factor, and when we get to #5^2 = 25#, we find that:

#175 = 25xx7#

So #sqrt175 = sqrt(25xx7) = sqrt25xxsqrt7=5sqrt7#.

We cannot further simplfy #sqrt7# so we'll simplify the other square root.

#28 = 4xx7#, so we get:

#sqrt28 = sqrt(4xx7) = 2sqrt7#

Here's the short way to write what we've done:

#sqrt175-sqrt28 = sqrt(25xx7)-sqrt(4xx7)#

# = 5sqrt7-2sqrt7#

Now, if we have 5 of these things and we subtract 2 of the things, surely we end up with 3 of the things.

So:

#sqrt175-sqrt28 = 5sqrt7-2sqrt7#

# = 3sqrt7#