Question

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 $a\ne 0)$ such that $\sin x+2\sin y=1$, then the value of $(a^2+2b^2+3ab+ac)$ equals

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $0$

Assume that $\sin \theta =x$. Hence, given quadratic is $a{x}^2+bx+c=0$ with roots $\sin x$ and $\sin y$.

$\therefore -\frac{b}{a}=\sin x+\sin y$ and $\frac{c}{a}=\sin x\sin y$

(sum and product of roots of quadratic equation)

$\therefore a^2+2b^2+3ab+ac=a^2(1+2{(-\frac{b}{a})}^2-3(-\frac{b}{a})+\left(\frac{c}{a}\right))$

$=a^2(1+2(\sin x+\sin y{)}^2-3(\sin x+\sin y)+\sin x\sin y)$

$=a^2(1+2{\sin }^{2}x+2{\sin }^{2}y+4\sin x\sin y-3\sin x-3\sin y+\sin x\sin y)$

$=a^2(1+2\sin x(\sin x+2\sin y)+\sin y(2\sin y+\sin x)-3\sin x-3\sin y)$

$=a^2(1+2\sin x+\sin y-3\sin x-3\sin y)$ $(\because \sin x+2\sin y=1)$

$=a^2(1-(\sin x+2\sin y\left)\right)$

$=a^2(1-1)$

$=0$

This is required answer.