Question

( (b, c, d) )
since ( x=2 ) and ( x=3 ) are roots of given equation ( therefore quad 3(2)^{2}-2 p(2)+2 q=0 )
( Rightarrow quad 12-4 p+2 q=0 Rightarrow-2 p+q=-6 )
and ( 3(3)^{2}-2 p(3)+2 q=0 )
( Rightarrow quad 27-6 p+2 q=0 Rightarrow-6 p+2 q=-27 )
On solving (i) and (ii), we get
( p=frac{15}{2} ) and ( q=-6+15=9 )

# If x= 2 and x = 3 are roots of the equation 3x2 - 2px + 2q = 0, then (a) p=3 C) 9=9 (b) pa 15 (d) bp-2q=27

Solution