if p,q,r are real numbers and quadratic equation px^2+qx+r=0 has no real root,then

(A)p(p+q+r)<0                                 (B)p(p-2q+4r)

(C)p(p+4q+2r)<0                            (D)none of these

Because as the eqn has no real roots , p and f(1), f(1/2) ,f(-1/2) must be of the same sign .

( f(x) = the LHS)

and f(1)= p +q+r

f(1/2)= (p+2q +4r)/4

f(-1/2)=(p-2q + 4r)/4 etc

so their product must be >0

Now writ option (B) clearly and chek the answer

otherwise it is (D)

Case 1 : -

if p < 0 for eqn having no real roots f (x) < 0 for all x for having non-real roots

hence f (1) < 0

( p + q + r ) < 0

hence we can say p (p + q + r) > 0

p < 0 hence f (-1 / 2) < 0

p (1 / 4) + q (-1 / 2 ) + r < 0

p - 2q + 4 r < 0

hence p (p - 2q + 4 r) > 0

if p < 0 hence for non-real roots f(1 / 2) < 0

p(1 / 4) + q (1 / 2) + r < 0

p + 2q +4 r < 0

hence p (p + 2q + 4r ) > 0

Case 2 : -

if p > 0 for eqn having no real roots f (x) > 0 for all x for having non-real roots

hence f (1) > 0

( p + q + r ) > 0

hence we can say p (p + q + r) > 0

p > 0 hence f (-1 / 2) > 0

p (1 / 4) + q (-1 / 2 ) + 4 r > 0

p - 2q + 4 r > 0

hence p (p - 2q + 4 r) > 0

if p > 0 hence for non-real roots f(1 / 2) > 0

p(1 / 4) + q (1 / 2) + r > 0

p + 2q +4 r > 0

hence p (p + 2q + 4r ) > 0