How do you simplify #sqrt48+sqrt147#?

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Answer 1 · 2018-05-29T16:04:14

See a solution process below:

First, we can rewrite the terms under the radicals as:

#sqrt(16 * 3) + sqrt(49 * 3)#

Now, we can use this rule of exponents to do the simplification:

#sqrt(color(red)(a) * color(blue)(b)) = sqrt(color(red)(a)) * sqrt(color(blue)(b))#

#(sqrt(16) * sqrt(3)) + (sqrt(49) * sqrt(3)) =>#

#4sqrt(3) + 7sqrt(3) =>#

#(4 + 7)sqrt(3) =>#

#11sqrt(3)#

Answer 2 · 2018-05-29T16:04:40

#11sqrt3#

#sqrt48 +sqrt147#

#sqrt16*sqrt3 +sqrt49*sqrt3#

#4sqrt3 +7sqrt3#

#11sqrt3#

#----------------#

You must find first number which should be a square number.

Like #16# and #49# are square numbers.

Answer 3 · 2018-05-29T16:07:39

#11sqrt3#

#sqrt48+sqrt147#

#sqrt(2*2*2*2*3)+sqrt(3*7*7)#

#sqrt(2^2*2^2*3)+sqrt(3*7^2)#

#2*2*sqrt(3)+7*sqrt(3)#

#4sqrt(3)+7sqrt(3)#

#11sqrt3#