If `|(a,b,aalpha+b),(b,c,balpha+c),(aalpha+b,balpha+c,0)|=0,` prove that a,b,c, are in G.P. or `alpha` is a root of `ax^(2)+`2bx+c=0

Options

A. a,b,c are in A.P
B. 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

B, D

Explanation

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`