How do you rationalize the denominator and simplify #1/(sqrt2-sqrt8)#?

Question

943 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T23:51:49

See a solution process below:

Multiply the fractions by this form of #1# to rationalize the denominator by removing the radicals from the denominator:

#(color(red)(sqrt(2)) + color(red)(sqrt(8)))/(color(red)(sqrt(2)) + color(red)(sqrt(8)))#

This gives:

#(color(red)(sqrt(2)) + color(red)(sqrt(8)))/(color(red)(sqrt(2)) + color(red)(sqrt(8))) xx 1/(sqrt(2) - sqrt(8)) =>#

#(sqrt(2) + sqrt(8))/((color(red)(sqrt(2)) + color(red)(sqrt(8))) xx (sqrt(2) - sqrt(8))) =>#

#(sqrt(2) + sqrt(8))/((color(red)(sqrt(2))sqrt(2) - color(red)(sqrt(2))sqrt(8) + color(red)(sqrt(8))sqrt(2) - color(red)(sqrt(8))sqrt(8))) =>#

#(sqrt(2) + sqrt(8))/(sqrt(2)^2 - color(red)(sqrt(2))sqrt(8) + sqrt(2)color(red)(sqrt(8)) - sqrt(8)^2) =>#

#(sqrt(2) + sqrt(8))/(2 - 0 - 8) =>#

#(sqrt(2) + sqrt(8))/(-6) =>#

#-(sqrt(2) + sqrt(8))/6#