Question

if sinA/2 and cosB/2 are the roots of px^2 + qx + r -0 and A= pie/3

p + 2q + 4r =0 p+ 2q = 4r

Answer 1

Put x = 1/2 in the equation: (Because sinA\2 = sin pi/6 = 1/2)


\[\frac{p}{4} + \frac{q}{2} + r = 0\]
\[p + 2q + 4r = 0\]

Answer 2

for p+ 2q = 4r
i think one of p and q must be 0