How do you rationalize the denominator and simplify #sqrt15/(sqrt15-sqrt13)#?

2 Answers
Mar 10, 2018

Answer:

See a solution process below:

Explanation:

To rationalize the denominator multiply the fraction by the appropriate form of #1#

#(color(red)(sqrt(15)) + color(red)(sqrt(13)))/(color(red)(sqrt(15)) + color(red)(sqrt(13))) xx sqrt(15)/(sqrt(15) - sqrt(13)) =>#

#(color(red)(sqrt(15))sqrt(15) + color(red)(sqrt(13))sqrt(15))/(color(red)(sqrt(15))sqrt(15) - color(red)(sqrt(15))sqrt(13) + color(red)(sqrt(13))sqrt(15) - color(red)(sqrt(13))sqrt(13)) =>#

#(15 + sqrt(color(red)(13) * 15))/(15 - 0 - 13) =>#

#(15 + sqrt(195))/2#

Mar 10, 2018

Answer:

#(15+sqrt195)/2#

Explanation:

#sqrt15/(sqrt15-sqrt13)#

#:.color(magenta)((sqrt15+sqrt13)/(sqrt15+sqrt13)=1#

#:.sqrt15/(sqrt15-sqrt13)xxcolor(magenta)((sqrt15+sqrt13)/(sqrt15+sqrt13)#

#:.color(magenta)(=sqrt15xxsqrt15=15#

#:.=(sqrt15(sqrt15+sqrt13))/((sqrt15-sqrt13)(sqrt15+sqrt13))#

#:.=(sqrt15(sqrt15+sqrt13))/(15-13)#

#:.=(sqrt15(sqrt15+sqrt13))/2#

#:.=(15+sqrt13sqrt15)/2#

#:.=(15+sqrt 195)/2#