How do you evaluate #3\sqrt { 45} + 3\sqrt { 5} + 2\sqrt { 27}#?

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Answer 1 · 2017-02-01T16:33:30

#=12sqrt5 +6sqrt3 = 37.225 " "#(3 d.p.)

#45, 5 and 27# are not square numbers and therefore do not have exact roots.

Write the numbers under the roots as the product of their prime factors, then you will know what you are working with.

#3sqrt45 +3sqrt5 +2sqrt27#

#=3sqrt((3xx3)xx5) + 3sqrt5 + 2sqrt((3xx3)xx3)#

Find square roots where you can.

#3xx3sqrt5 +3sqrt5 + 2xx3sqrt3#

#=9sqrt5 +3sqrt5+6sqrt3" "larr# add like terms.

#=12sqrt5 +6sqrt3#

To find an actual value you will need a calculator.

Answer 2 · 2017-02-01T17:13:26

#6(2sqrt5+sqrt3) or 37.225#

#3sqrt45+3sqrt5+2sqrt27#

#:.=3sqrt(3*3*5)+3sqrt5+2sqrt(3*3*3#

#:.=3*sqrt3*sqrt3*sqrt5+3sqrt5+2*sqrt3*sqrt3*sqrt3#

#:.=3*3*sqrt5+3sqrt5+2*3sqrt3#

#:.=9*sqrt5+3sqrt5+6*sqrt3#

#:.=9sqrt5+3sqrt5+6sqrt3#

#:.=12sqrt5+6sqrt3#

#:.=6(2sqrt5+sqrt3)# or #37.225#