How do you simplify square root of 128 + the square root of 32?

4 Answers
May 2, 2018

#sqrt(128)+sqrt(32)#

#sqrt((4 xx 4) xx (2 xx 2) xx 2) + sqrt((4 xx 4) xx 2)#

#4 xx 2sqrt(2) + 4sqrt(2)#

#8sqrt(2) + 4sqrt(2)#

#12sqrt(2)#

May 2, 2018

Answer:

#12sqrt2#

Explanation:

Simplify:

#sqrt128=sqrt16xxsqrt8 -> 4 xx sqrt4 xx sqrt2=8sqrt2#.

#sqrt32=sqrt16xxsqrt2=4sqrt2#

#8sqrt2+4sqrt2=12sqrt2#

May 2, 2018

Answer:

#\sqrt(128)+\sqrt(32)=2*\sqrt(12)#

Explanation:

#\sqrt(128)+\sqrt(32)#
Write both numbers as a product of prime numbers.
(unique factorisation ).

For example:
my pic

#128=2*64=2*2*32#
#32=2*16=2*2*8=2*2*2*4=2^5=2^4*2#

#\sqrt(128)+\sqrt(32)=\sqrt(2*2*32)+\sqrt(32)=#
#=\sqrt(2*2)* \sqrt(32)+\sqrt(32)=2* \sqrt(32)+\sqrt(32)=#
#=3*\sqrt(32)=3* \sqrt(2^4*2)= 3*\sqrt(2^4)*\sqrt(2)= 3*2^2*\sqrt(2)=#
#=3*4*\sqrt(2)=12\sqrt(2)#

May 2, 2018

Answer:

#12sqrt(2)#

Explanation:

Given: #sqrt(128)+sqrt(32)#

You are looking for squared values that you can 'extract' from the roots.

If you are ever not sure do a quick sketch of a prime factor tree to help:
Tony B

#sqrt(128)+sqrt(32)#

#sqrt(2^2xx2^2xx2^2xx2)+sqrt(2^2xx2^2xx2)#

#color(white)("ddddd")8sqrt(2)color(white)("dddddd")+color(white)("dddd")4sqrt(2)color(white)("dddd")=color(white)("d")12sqrt(2)#