English

Factorise : (9a)^2 + 1/(9a)^2 - 2 - 12a + 4/(3a) - Mathematics

Advertisements
Advertisements

Question

Factorise : `(9a)^2 + 1/(9a)^2 - 2 - 12a + 4/(3a)`

Sum

Solution

`(9a)^2 + 1/(9a)^2 - 2 - 12a + 4/(3a)`

= `(3a)^2 + 1/(3a)^2 - 2 xx 3a xx 1/(3a) - 4( 3a - 1/(3a))`

= `( 3a - 1/(3a))^2 - 4( 3a - 1/(3a))`

= `(3a - 1/(3a))[( 3a - 1/(3a)) - 4]`

= `( 3a - 1/(3a))( 3a - 4 - 1/(3a))`

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Factorisation - Exercise 5 (E) [Page 76]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 5 Factorisation
Exercise 5 (E) | Q 2 | Page 76
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×