How do you simplify #( sqrt32)/ (5 sqrt 14)#?

2 Answers
Feb 26, 2016

Answer:

# (4sqrt(7))/35#

Explanation:

You have to try and spot common values that can be cancelled out.

Both 32 and 14 are even so 2 has to be a common factor giving:

#" "(sqrt(2xx16))/(5sqrt(2xx7))#

7 is a prime number so the denominator can not be broken down any further

However, 16 in the numerator can be broken down into #4^2# so we now have:

#" "(sqrt(2xx4^2))/(5sqrt(2xx7))#

#" "1/5xxsqrt(2)/(sqrt(2)) xxsqrt(4^2)/sqrt(7)#

#" " 1/5xx 1 xx 4/sqrt(7)" "=" "4/(5sqrt(7))#

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Added extension to solution
I forgot that standard practice is that you try to remove any roots from the denominator!

Multiply by 1 but in the form of #sqrt(7)/sqrt(7)# giving:

#" "4/(5sqrt(7))xx sqrt(7)/sqrt(7)" "=" "(4sqrt(7))/(5xx7)" "=" "(4sqrt(7))/35#

Feb 26, 2016

Answer:

#sqrt 32/(5sqrt 14)=(4sqrt7)/35#

Explanation:

#sqrt 32/(5sqrt 14)#

Simplify #sqrt 32#.

#sqrt(2xx2xx2xx2xx2)=#

#sqrt(2^2xx2^2xx2)=#

#4sqrt2#

Add this back into the expression.

#(4sqrt 2)/(5sqrt14)#

Rationalize the denominator by multiplying both the numerator and denominator by #sqrt 14#.

#(4sqrt 2)/(5sqrt14)xx(sqrt14)/sqrt 14=#

#(4sqrt 2sqrt14)/(5xx14)#

Simplify the numerator by multiplying the square roots.

#(4sqrt 28)/(5xx14)#

Simplify the square root by factoring.

#(4sqrt 28)/(5xx14)=(4sqrt(2xx2xx7))/(5xx14)=(4xx2sqrt7)/(5xx14)=(8sqrt7)/(5xx14)#

Simplify the denominator.

#(8sqrt 7)/(70)#

Simplify the expression.

#(4sqrt7)/35#