How do you rationalize the denominator and simplify #1/(1-8sqrt2)#?

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Answer 1 · 2018-03-25T03:10:45

I believe this should be simplified as #(-(8sqrt2+1))/127#.

To rationalize the denominator, you must multiply the term that has the #sqrt# by itself, to move it to the numerator. So:

#=>##1/(1-8*sqrt2)*8sqrt2#

This will give:

#=>##(8sqrt2+1)/(1-(8sqrt2)^2#

#(8sqrt2)^2=64*2=128#

#=>##(8sqrt2+1)/(1-128)#

#=>##(8sqrt2+1)/-127#

The negative cam also be moved to the top, for:

#=>##(-(8sqrt2+1))/127#

Answer 2 · 2018-03-25T03:14:30

#(-1-8sqrt2)/127#

Multiply the numerator and the denominator by the surd (to undo the surd) and the negative of the extra value.

#1/(1-8sqrt2# x #(-1+8sqrt2)/(-1+8sqrt2#

#(1(1+8sqrt2))/((1-8sqrt2)(1+8sqrt2)#

Expand brackets. Use the FOIL rule for the denominator.

#(1+8sqrt2)/-127#

You could simplify further by taking the negative of the denominator and apply it to the numerator.

#(-1-8sqrt2)/127#