How do you simplify #6sqrt7+2sqrt28#?

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Answer 1 · 2018-05-14T19:14:35

#=>10sqrt(7)#

We are given

#6sqrt7 + 2sqrt(28)#

We can factor the #28# to find a perfect square that can then be pulled out of the radical.

#=6sqrt7 + 2sqrt(4*7)#

#=6sqrt7 + 2sqrt(2^2 * 7)#

#=6sqrt7 +2*2sqrt(7)#

#=6sqrt7 + 4sqrt(7)#

Since the radicals are the same, we can combine like-terms using distribution.

#=(6+4)sqrt(7)#

#=10sqrt(7)#

Answer 2 · 2018-05-14T19:16:25

26.45751311065

#6sqrt(7)# + #2sqrt(28)#

First, let's unsimplify these terms in order to make 'em easier to combine. Any number that is outside the square root has a mate.

So, the 6 outside of #sqrt(7)# is actually 6 * 6, which is then also multiplied by 7. So:

#6sqrt(7)# becomes the square root of #6 * 6 * 7#, which is #sqrt(252)#. To double check, they should be the same, like this:

#6sqrt(7)# = 15.87450786639
#sqrt(252)# = 15.87450786639

Do the same with your other square root. #2sqrt(28)# is actually #2 * 2# multiplied by 28. So:

#2sqrt(28)# becomes the square root of #2 * 2 * 28#, which is: #sqrt(112)#. To double check:

#2sqrt(28)# = 10.58300524426
#sqrt(112)# = 10.58300524426

Now, add your two unsimplified square roots:

#sqrt(112)# + #sqrt(252)# = 26.45751311065