If `a+b+c=0,` one root of `|a-x c b c b-x a b a c-x|=0` is `x=1` b. `x=2` c. `x=a^2+b^2+c^2` d. `x=0`

Options

A. x = a
B. `x = sqrt((3(a^(2)+b^(2)+c^(2)))/(2))`
C. `x = sqrt((2(a^(2)+b^(2)+c^(2)))/(3))`
D. x = 0

Answer

D

Explanation

`|{:(a-x," "c," "b),(" "c,b-x," "a),(" "b," "a,c-x):}|=0`
`-|{:(x-(a+b+c)," "c," "b),(x-(a+b+c),b-x," "a),(x-(a+b+c)," "a,c-x):}|=0`
`rArr |{:(x," "c," "b),(x,b-x," "a),(x," "a,c-x):}|=0`
{Applying `C_(1) rarr C_(1) + C_(2) + C_(3)`}
`|{:(x," "c," "b),(0,c+x-b," "b-a),(0," "c-a,b+x-c):}|=0`
{Applying `R_(2) rarr R_(1) - R_(2) and R_(3) rarr R_(1) - R_(3)`}
`x{(c+x-b)(b+x-c)-(b-a)(c-a)}=0`
` x{(x^(2)-(b-c)^(2)-bc + ac + ab - a^(2)}=0`
`x(x^(2)-a^(2)-b^(2)-c^(2)+ab+bc+ca)=0`
`x{x^(2)-(a-b)^(2)-(b-c)^(2)-(c-a)^(2)}=0`
`therefore x = 0`