How do you simplify #sqrt(49)/sqrt(500)#?

1 Answer
Jan 29, 2016

#sqrt(49)/sqrt(500)=7/(10sqrt(5))=(7sqrt(5))/50#

Explanation:

We will use the following properties:

  • If #a >= 0# then #sqrt(a^2) = a#

  • If #a >=0# or #b >=0# then #sqrt(ab) = sqrt(a)sqrt(b)#

#sqrt(49)/sqrt(500) = sqrt(7^2)/sqrt(10^2*5)#

#=7/(sqrt(10^2)sqrt(5))#

#=7/(10sqrt(5))#

We could finish here, or rationalize the denominator by multiplying the numerator and denominator by #sqrt(5)# to obtain

#7/(10sqrt(5)) = (7sqrt(5))/(10sqrt(5)sqrt(5))#

#=(7sqrt(5))/(10*5)#

#=(7sqrt(5))/50#