Question

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

Answer 1

`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)`

Answer 2

`12sqrt2`

Explanation:

Simplify:

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

`sqrt32=sqrt16xxsqrt2=4sqrt2`

`8sqrt2+4sqrt2=12sqrt2`

Answer 3

`\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:

`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)`

Answer 4

`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:

`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)`