(2+√52−√5)+(2−√52+√5)
It can be solved by rationalizing the denominators individually. For that, we first find the rationalizing factor for both the denominators, (2−√5) and (2+√5). [1 Mark]
The rationalizing factor for
(2−√5) would be (2+√5) and that for (2+√5) would be (2−√5).
Rationalizing the given term we get,
(2+√5)(2+√5)(2−√5)(2+√5)+(2−√5)(2−√5)(2+√5)(2−√5) [1 Mark]
=4+5+4√54−5+4+5−4√54−5
=4+5+4√5+4+5−4√5−1
= -18 [1 Mark]