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

Question

Abigail Allen · Accepted Answer

#4 - 2 sqrt 2# 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#

Isaiah Anthony · Answer

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)#

Savannah Anderson · Answer

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

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)#

Eva Adamson · Answer

undefined To simplify the expression sqrt(16) / (sqrt(4) + sqrt(2)), we can start by simplifying the square roots.

The square root of 16 is 4, and the square root of 4 is 2.

Therefore, the expression becomes 4 / (2 + sqrt(2)).

To simplify further, we can rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is 2 - sqrt(2).

This gives us (4 * (2 - sqrt(2))) / ((2 + sqrt(2)) * (2 - sqrt(2))).

Simplifying the denominator further, we have (4 * (2 - sqrt(2))) / (4 - 2).

Finally, simplifying the numerator, we get (8 - 4sqrt(2)) / 2.

This can be further simplified to 4 - 2sqrt(2).

Therefore, the simplified expression is 4 - 2sqrt(2).