Question

Let the roots be ( x ) and ( x^{2} )
By product of roots ( x^{3}=1 )
( x^{3}-1=0 )
( (x-1)left(x^{2}-x-1right)=0 )
which implies ( x=1 ) or ( x^{2}-x=-1 ) If we take ( x=1 ) then by sum of roots ( p=-6 ) (but it is rejected as ( p>0 ) ) If we take ( x^{2}-x=-1 ) then by equating with sum of roots ( -1=-p / 3 ) which implies ( p=3 )

# For the equation 3x + px + 3-0. p > O if one of the roots is square of the other, then pis equal to (A) 1/3 (B) 1 (C) 3 (D) 2/3

Solution