Question

( alpha+beta=-beta )
Ifiron ( quad ) and ( x^{3}+3^{3}=q )
( begin{aligned} x^{3}+beta^{3}=&(x+beta)left(x^{2}+beta^{2}-x betaright) &=(alpha+beta)left[(alpha+beta)^{2}-3 alpha betaright) end{aligned} )
( left.therefore quad(alpha+beta)left[(alpha+beta)^{2}-3 alpha betaright)right]=q )
( left.Rightarrow-pleft[(-p)^{2}-3 alpha betaright)right]=q )
( Rightarrow quad 3 alpha beta-p^{2}=frac{q}{p} )
( Rightarrow quad alpha beta=frac{1}{3}left(frac{q}{p}+phi^{2}right) )
( e q^{n} ) with roota ( x / beta ) and ( frac{beta}{alpha} ) be ( x^{2}+b x+c=0 )
( frac{alpha}{R}+frac{beta}{alpha}=-b )
( therefore quad operatorname{eqn} i s quad x^{2}-left(frac{b^{3}-2 q}{p^{3}+q}right) x+1=0 )
Hene ans ( { }^{circ} mathrm{s}(underline{B}) )

# UD uUL 15 UTUL et pand q be real numbers such that p + 0.03 # g and p3 # -9. a numbers satisfying a + B = -p and a3 + b3 = q, then a quadratic equation po + q and p3 + - q. If a and B are nonzero complex ation having a and as its roots [IIT-JEE 2010, Paper-1,(3,-1)/ 84] (B) (p3 +q) x2 - (p3-29)+ (p3 + 9) = 0 (D) (93-9) x2 - (5p3 + 2q)x + (p2 - 9) = 0 (A) (p3 + q) x2 - (p3 + 2q)x + (p3 + q) = 0 c) (p3 - q) x2 - (5p3 - 2q)X + (p3 - 9) = 0 unen for n > 1, then the value of

Solution