How do you simplify #sqrt (16) / (sqrt (4) + sqrt (2))#?

3 Answers

It is

#sqrt (16) / (sqrt (4) + sqrt (2))=4/[sqrt2(sqrt2+1)]= 2sqrt2/(sqrt2+1)=2*sqrt2(sqrt2-1)/[(sqrt2+1)*(sqrt2-1)]= 2sqrt2(sqrt2-1)#

Jun 26, 2016

Answer:

#4 - 2 sqrt 2#

Explanation:

Try to rationalize the denominator. Multiply numerator and denominatr by # (sqrt 4 - sqrt 2)#

#sqrt 16 ( sqrt 4 - sqrt 2) / ( (sqrt 4+ sqrt 2 ) * (sqrt 4 - sqrt 2 ) )#

#4 * (2 - sqrt 2) / ( 4 - 2) #

#4 * (2 - sqrt 2) / 2 #

#2 * (2 - sqrt 2) #

# 4 - 2sqrt2#

Answer:

Multiply through by #(2-sqrt2)/(2-sqrt2)# and work through to get #4-2sqrt2=2(2-sqrt2)#

Explanation:

Let's start with the original:

#sqrt16/(sqrt4+sqrt2)#

Let's first take the square roots of the perfect squares:

#4/(2+sqrt2)#

In order to simplify, we need the square root out from the denominator. The way to do this is to ensure that when we do FOIL (the process of multiplying 2 quantities within brackets), we don't end up with more square roots. To do that, we'll multiply by #(2-sqrt2)# which will eliminate that possibility (like this):

#4/(2+sqrt2)*((2-sqrt2)/(2-sqrt2))#

#(4*2-4sqrt2)/(2*2-2sqrt2+2sqrt2-sqrt2sqrt2)#

#(8-4sqrt2)/(4-2)=(8-4sqrt2)/2=4-2sqrt2=2(2-sqrt2)#