Let a, be the roots of the equation `x^2+x+1=0`. The equation whose roots are `alpha^19` and `beta^7` are:

Options

A. `x^(2) - x- 1 = 0`
B. `x^(2) - x+ 1=0`
C. `x^(2) + x-1 = 0`
D. `x^(2) + x+ 1 = 0`

Answer

B

Explanation

Given Equation is `x^(2) + x + 1 = 0`
`rArr x= (-1pm sqrt(3))/2`
Thus ,roots are `omega and omega ^(2)`
`:. alpha = omega and beta = omega^(2)`
So , `alpha^(19) = (omega)^(19)=(omega^(3))^(6).omega = omega " " ( :' omega^(3) =1)`
and, `alpha^(19) + (omega^(2))^(7)=omega^(14) = (omega^(3))^(4).omega^(2). omega^(2)=omega^(2)`
Now, `alpha^(19) + beta^(7)=omega + omega^(2) = -1 ( :'1+omega + omega^(2) =0)`
and `(alpha^(19))(beta^(7)) = (omega) (omega^(2))=1`
`:. " Required Quadratic equation whose roots are "alpha^(19)and beta^(7)` is
`x^(2) + (alpha^(19)+ beta^(7)) x+(alpha^(19))(beta^(7))=0`
`rArr x^(2) + (-1) x + 1=0`
`rArr x^(2) - x+1 = 0`