What is #sqrt72 + sqrt18#?

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Answer 1 · 2017-05-03T13:58:47

See the solution process below:

We can use this rule of radicals to rewrite this expression:

#sqrt(a * b) = sqrt(a) * sqrt(b)#

#sqrt(72) + sqrt(18) = sqrt(4 * 18) + sqrt(18) = (sqrt(4) * sqrt(18)) + sqrt(18) =#

#+-2sqrt(18) + sqrt(18) = +-2sqrt(18) + 1sqrt(18) = (+-2 + 1)sqrt(18)#

We can apply the rule again to #sqrt(18)#:

#(+-2 + 1)sqrt(18) = (+-2 + 1)sqrt(9 * 2) = (+-2 + 1)(sqrt(9) * sqrt(2)) =#

#(+-2 + 1)(+-3sqrt(2))#

For: #(+2 + 1) = 3#

#3(+-3sqrt(2)) = +-9sqrt(2)#

For: #(-2 + 1) = -1#

#-1(+-3sqrt(2)) = +-3sqrt(2)#

The solutions are:

#+-9sqrt(2)# or #+-3sqrt(2)#