Let $\sin x$ and $\sin y$ be roots of the quadratic equation $a\sin^2\theta+b\sin \theta+c=0\, (a, b, c \in R$ and $a \neq 0)$ such that $\sin x+2\sin y=1$, then the value of $(a^2+2b^2+3ab+ac)$ equals

Let the roots be r and s, so r + 2s = 1. Substituting r = 1 - 2s into the given relation 4rs = r^2 + s^2 gives 4(1 - 2s)s = (1 - 2s)^2 + s^2, which simplifies to 5s^2 - 4s = 0, meaning s = 4/5 and r = -3/5. For the original equation, the sum of the roots is -b/a = 1/5 and the product is c/a = -12/25. Substituting these values into the expression a^2 + 2b^2 + 3ab + ac gives a^2(1 + 2(1/25) + 3(-1/5) - 12/25) = a^2(1 + 2/25 - 15/25 - 12/25) = 0.

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