Question

If the roots of $a{x}^2+bx+c$are $\sin \alpha$ and $\cos \beta$ for some $\alpha$ then which one of the following is correct?

• A. a² + b² = 2ac
• B. b² – c² = 2ab
• C. b² – a² = 2ac
• D. b² + c² = 2ab

Answer 1

The correct option is C

b² – a² = 2ac

Given:

$\sin \alpha \&\cos \beta$ are roots for some $\alpha$ of the above equation $a{x}^2+bx+c=0$

Step-1 Find the sum and product of roots of $a{x}^2+bx+c=0$

$\sin \alpha \&\cos \beta$ are roots for some $\alpha$ of the above equation.

Sum of the roots = - coefficient of $x$ / coefficient of $x^2$

$\sin \alpha +\cos \beta =\frac{-b}{a}$ …(1)

Product of roots = Constant / coefficient of $x^2$

$\sin \alpha \times \cos \beta =\frac{c}{a}$…(2)

Step-2 Solve equations (1) and (2)

Squaring both sides of equation (1)

$$\begin{gathered} \Rightarrow {(\sin \alpha +\cos \beta )}^2=(\frac{-b}{a}{)}^2 \\ \Rightarrow {\sin }^2\alpha +{\cos }^2\beta +2\sin \alpha \cos \beta =\frac{b^2}{a^2} \end{gathered}$$

Use the trigonometric Identity ${\sin }^2+{\cos }^{2}x=1$ and equation (2).

$1+2\frac{c}{a}=\frac{b^2}{a^2}$

Subtracting both sides by 1.

$$\begin{gathered} \Rightarrow 1+2\frac{c}{a}-1=\frac{b^2}{a^2}-1 \\ \Rightarrow 2\frac{c}{a}=\frac{b^2-a^2}{a^2} \\ \Rightarrow 2ac=b^2-a^2 \\ \Rightarrow b^2-a^2=2ac \end{gathered}$$

Therefore option (C) $b^2-a^2=2ac$ is the correct option.