How do you rationalize the denominator and simplify #sqrt(10/7)#?

2 Answers
Jul 9, 2017

Answer:

#sqrt(70)/7#

Explanation:

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

To rationalise this denominatir, we multiply the top and bottom by #sqrt(7)#, so we get #sqrt(10)/sqrt(7)*sqrt(7)/sqrt(7)#, #sqrt(7)*sqrt(7)# is just #7#, and #sqrt(10)*sqrt(7)=sqrt(10*7)=sqrt(70)#, as #70# doesn't have any factors which are perfect squares, it can only stay as #sqrt(70)#, giving us #sqrt(70)/7#

Jul 9, 2017

Answer:

See the solution process below:

Explanation:

First, we need to rewrite the expression using this rule for dividing radicals:

#sqrt(color(red)(a)/color(blue)(b)) = sqrt(color(red)(a))/sqrt(color(blue)(b))#

#sqrt(color(red)(10)/color(blue)(7)) = sqrt(color(red)(10))/sqrt(color(blue)(7))#

Next, we can rationalize the denominator and remove the radical by multiplying the fraction by the appropriate form of #1#:

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

We can now use this rule of exponents to simplify the numerator:

#sqrt(color(red)(a)) * sqrt(color(blue)(b)) = sqrt(color(red)(a) * color(blue)(b))#

#(sqrt(color(red)(7)) * sqrt(color(blue)(10)))/7 = sqrt(color(red)(7) * color(blue)(10))/7 = sqrt(70)/7#