How do you evaluate #2root5{ 2} + 2root5{ 64}#?

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Answer 1 · 2017-08-04T04:13:30

See a solution process below:

First, we can rewrite the expression as:

#2root(5)(2) + 2root(5)(32 * 2) =>#

#2root(5)(2) + 2root(5)(2^5 * 2)#

Now we can use this rule of radicals to simplify the radical on the right side of the expression:

#root(n)(color(red)(a) * color(blue)(b)) = root(n)(color(red)(a)) * root(n)(color(blue)(b))#

#2root(5)(2) + 2root(5)(color(red)(2^5) * color(blue)(2)) =>#

#2root(5)(2) + 2root(5)(color(red)(2^5))root(5)(color(blue)(2)) =>#

#2root(5)(2) + (2 * 2root(5)(color(blue)(2))) =>#

#2root(5)(2) + 4root(5)(color(blue)(2))#

We can now factor out the common term of #root(5)(3)# and combine the remaining terms:

#2root(5)(2) + 4root(5)(2) =>#

#(2 + 4)root(5)(2) =>#

#6root(5)(2)#