How do you simplify #4 sqrt 3 - sqrt 64 + 6 sqrt 27#?

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Answer 1 · 2015-11-07T23:27:45

#22sqrt3−8#

We know that #sqrt 64 = 8#, so we have

#4sqrt3−8+6sqrt27#

Now let's rewrite #sqrt27# as #sqrt(9)*sqrt3#
since we know the square root of 9, we can write it as #3sqrt3#

Now just a bit of simple multiplication and combining link terms
#4sqrt3−8+6(3sqrt3)#

#4sqrt3−8+18sqrt3)#

#22sqrt3−8#

Answer 2 · 2015-11-07T23:41:10

22sqrt[3] -8

We can simplify first by solving sqrt[64] since we know it is a perfect square. This leaves:

4sqrt[3] -8 + 6sqrt[27]

Then we need to see if we can simplify the square roots.

4sqrt[3] is as simplified as it gets.
6sqrt[27] can be broken up into factors so:

6sqrt[27] = 6sqrt[3*9]

9 is a perfect square so it can be rooted and brought out side of the square root. This leaves:

6(3)sqrt[3]

So now we have:

4sqrt[3] -8 + 18sqrt[3]

Combine terms:

22sqrt[3] - 8