the determinant `|{:(a,,b,,aalpha+b),(b,,c,,balpha+c),(aalpha+b,,balpha+c,,0):}|=0` is equal to zero ifA. a,b,c are in A.PB. a,b,c are in G.P.C. `alpha` is a root of the equation `ax^(2) +bx+c=0`D. `(x-alpha)` is a factor fo `ax^(2) +2bx+c`

the determinant `|{:(a,,b,,aalpha+b),(b,,c,,balpha+c),(aalpha+b,,balph

1.	the determinant `\|{:(a,,b,,aalpha+b),(b,,c,,balpha+c),(aalpha+b,,balpha+c,,0):}\|=0` is equal to zero ifA. a,b,c are in A.PB. a,b,c are in G.P.C. `alpha` is a root of the equation `ax^(2) +bx+c=0`D. `(x-alpha)` is a factor fo `ax^(2) +2bx+c`
Answer» Correct Answer - B::D Given that `\|{:(a,,b,,aalpha+b),(b,,c,,balpha+c),(aalpha+b,,balpha+c,,0):}\|=0` Operating `C_(3) to C_(3) -C_(1) alpha-C_(2)` we get `\|{:(a,,b,,0),(b,,c,,0),(aalpha+b,,balpha+c,,-(aalpha^(2)+balpha+balpha+c)):}\|=0` `rArr (aalpha^(2) +2balpha +c) \|{:(a,,b,,0),(b,,c,,0),(aalpha +b,,balpha+c,,1):}\|=0` `rArr (ac-b^(2)) (aalpha^(2) +2balpha+c)=0` so , either `ac-b^(2)=0 " or " aalpha^(2) +2balpha +c=0` This means that either a,b,c are in G.P. or `(x-alpha)` is a factor of `ax^(2) +2bx +C`