यदि `a = 9 - 4sqrt(5)`, तो `sqrt(a) - 1/(sqrt(a))` का मान ज्ञात कीजिए ।

Explanation

यहाँ `a = 9 - 4sqrt(3)"….."(1)`
`:. 1/a = 1/(9-4sqrt(5)) = 1/(9-4sqrt(5)) = (1)/(9-4sqrt(5)) xx (9+4sqrt(5))/(9+4sqrt(5)) = (9+4sqrt(5))/((9)^(2) -(4sqrt(5))^(2)) = (9+4sqrt(5))/(81-80) = 9 + 4sqrt(5)"…."(2)`
अब `(sqrt(a) - 1/(sqrt(a)))^(2) = a+ 1/a - 2sqrt(a).(1)/(sqrt(a)) = a + 1/a - 2 = (9-4sqrt(5)) + (9+4sqrt(5)) = 2`
`rArr (sqrt(a) - 1/(sqrt(a)))^(2) = 16`
दोनों पक्षों का वर्ग करने पर
`:. sqrt(a) - 1/(sqrt(a)) = +=sqrt(16) rArr sqrt(a) - 1/(sqrt(a)) = +-4`
परन्तु `a lt 1/a rArr sqrt(a) lt 1/(sqrt(a))`
`rArr sqrt(a) - 1/(sqrt(a)) lt 0`
`:. sqrt(a) - 1/(sqrt(a)) = -4`