How would you simplify #sqrt48 + sqrt3#?

4 Answers
Feb 13, 2016

Answer:

#5sqrt3#

Explanation:

We can split up #sqrt48# using the rule that #sqrt(ab)=sqrtasqrtb#.

Thus, #sqrt48=sqrt(16*3)=sqrt16sqrt3=4sqrt3#.

The original expression can be rewritten as

#4sqrt3+sqrt3=sqrt3(4+1)=5sqrt3#

Feb 13, 2016

Answer:

Short story, try watching the result of division of the two numbers and substitute it in the bigger one.

#sqrt(48)+sqrt(3)=5sqrt(3)#

Explanation:

Always try reducing the high number to something of which you know the root. Here someone should notice that:

#48/3=16<=>48=3*16#

The root of 16 is known, while the root of 3 can be factored. Therefore:

#sqrt(48)+sqrt(3)#

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

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

#4*sqrt(3)+sqrt(3)#

#5sqrt(3)#

Feb 13, 2016

Answer:

#5sqrt3 #

Explanation:

radicals in simplified form are #asqrtb #
where a is a rational number.

to begin with , simplify # sqrt48 #

by considering the factors of 48 , particularly 'squares'

the factors required here are 16 (square) and 3.

using the following : # sqrta xx sqrtb hArr sqrtab #

#sqrt48 = sqrt16 xx sqrt3 = 4sqrt3 #

hence #sqrt48 + sqrt3 = 4sqrt3 + sqrt3 = 5sqrt3 #

Feb 13, 2016

Answer:

#color(blue)(5sqrt3)#

:)

Explanation:

To simplify radicals, we must find first its largest "perfect squares" that evenly divides to simplify.

Since, #sqrt(ab) = sqrta*sqrtb# (Theorem)

#sqrt48 + sqrt 3#

PerfectSquares:

#color(blue)(4)# #= 2*2#
#color(blue)(9)# #= 3*3#
#color(blue)(16)# #= 4*4#
#color(blue)(25)# #= 5*5#
#color(blue)(36)# #= 6*6#
#color(blue)(49)# #= 7*7#
#color(blue)(64)# #= 8*8#
#color(blue)(81)# #= 9*9#
#color(blue)(100)# #= 10*10#

we can simplify #sqrt48#, as 48 evenly divides with #16#,

we get,

#sqrt48 = sqrt(16*3)#

using theorem from above,

#sqrt48 = sqrt(16)*sqrt(3)#

simplify,

#sqrt48 = 4*sqrt3#

#=4sqrt3#

since the rule of radical signs is just like variables we can combine like terms, with a slight difference.

#asqrtb+-asqrtb=(a+a)sqrtb#

#4sqrt3## +# #sqrt3#

#4sqrt3## +# #(1)##sqrt3#

#color(blue)(= 5sqrt3)#