What is the square root of 3 + the square root of 72 - the square root of 128 +the square root of 108?

1 Answer
Oct 1, 2015

Answer:

#7sqrt(3) - 2sqrt(2)#

Explanation:

#sqrt(3) + sqrt(72) - sqrt(128) + sqrt(108)#

We know that #108 = 9 * 12 = 3^3 * 2^2#, so #sqrt(108) = sqrt(3^3*2^2) = 6sqrt(3)#

#sqrt(3) + sqrt(72) - sqrt(128) + 6sqrt(3)#

We know that #72 = 9*8 = 3^2*2^3#, so #sqrt(72) = sqrt(3^2*2^3) = 6sqrt(2)#

#sqrt(3) + 6sqrt(2) - sqrt(128) + 6sqrt(3)#

We know that #128 = 2^7#, so #sqrt(128) = sqrt(2^6*2) = 8sqrt(2)#

#sqrt(3) + 6sqrt(2) - 8sqrt(2) + 6sqrt(3)#

Simplifying

#7sqrt(3) - 2sqrt(2)#