How do you evaluate #2\sqrt { 24} - \sqrt { 3} - 2\sqrt { 27}#?

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Answer 1 · 2017-05-04T20:34:36

#4sqrt6-7sqrt3#

expand all numbers:

#2sqrt24= sqrt4 * sqrt24#

#=sqrt(4*24)#

#=sqrt96#

#2sqrt27 = sqrt4 * sqrt27#

#=sqrt(4*27)#

#=sqrt108#

#2sqrt24 - sqrt3 - 2sqrt27 = sqrt96 - sqrt3 - sqrt108#

find all numbers as multiples of #sqrt3#:

#sqrt96 = sqrt3 * sqrt32 = sqrt3 * 2sqrt8# or #(sqrt32)(sqrt3)#

#sqrt108 = sqrt3 * sqrt36 = 6sqrt3#

#2sqrt24 - sqrt3 - 2sqrt27 = (sqrt32)(sqrt3) - sqrt3 - 6sqrt3#

#=(sqrt32-1-6)(sqrt3) = (sqrt32-7)(sqrt3) or (4sqrt2-7)(sqrt3)#

multiplied out, this is:

#4sqrt6-7sqrt3#

Answer 2 · 2017-05-13T01:20:01

Shorter way to get #4sqrt(6)-7sqrt(3)#

Break down the values inside the square root into their smallest components:

#2sqrt(2xx2xx2xx3) -sqrt(3)-2sqrt(3xx3xx3)#

If there are two of the same numbers, then you can take them out of the square root. For example, #sqrt(2xx2)# can be taken out as #2#

#2(2)sqrt(2xx3)-sqrt(3)-2(3)sqrt(3)#

Clean up the problem by multiplying everything together:

#4sqrt(6)-sqrt(3)-6sqrt(3)#

The #-sqrt(3)# and #-6sqrt(3)# can be combined by adding the coefficients together:

#-1+(-6)=-7#

So we have:

#4sqrt(6)-7sqrt(3)#