Question

rationalising the denominator of each of the following 1 by root 7 + root 6 subtraction root 13

Answer 1

Hey friend,

Here is your answer,

1/(√7+√6-√13)
Rationalizing factor :√7+√6+√13
=1/(√7+√6-√13)×
√7+√6+√13/(√7+√6+√13)

=1/(√7+√6+√13)/(√7+√6-√13)
(√7+√6+√13)

=√7+√6+√13/(√7+√6)²-(√13)²

=√7+√6+√13/[√7²+√6²+2(√7)(√6)]-13

=√7+√6+√13/13+2√42-13

=√7+√6+√13/2√42

The denominator is still irrational.
So we have to rationalize it further.

Now rationalizing factor: √42

=√7+√6+√13/2√42×√42/√42

=√42(√7+√6+√13)/2(√42)²

=√42×7+√42×6+√42×13/2(42)

=(7√6+6√7+√546)/84

=7√6/84+6√7/84+√546/84

=√6/12+√7/14+√546/84

Now the denominator is rationalised.