Question #f2e51

1 Answer
Mar 24, 2017

Answer:

#(5sqrt(2))/16#

Explanation:

You are looking for ways to split the numbers into squared values that you can 'take outside' the root

For example #25 ->5^2# so #sqrt(25)=sqrt(5^2)=5#

You can write #sqrt(25/128)# as #sqrt(25)/sqrt(128)#

If you are ever not sure about roots build a factor tree diagram (sketch) at the side of your work area. Do not forget to label it as rough work.

Tony B

#sqrt(25/128)" "= "sqrt(5^2)/(sqrt(2^2xx2^2xx2^2xx2))#

#=5/(2xx2xx2xxsqrt(2)) = 5/(8sqrt(2))#

Mathematicians do not like roots in the denominator so lets get rid of it.

Multiply by 1 and you do not change the intrinsic value.

#color(green)(5/(8sqrt(2))color(red)(xx1)" "->" "5/(8sqrt(2))color(red)(xxsqrt(2)/sqrt(2))#

#" "color(green)((5sqrt(2))/(8xx2)" "=" "(5sqrt(2))/16)#