Question

If $\sin \alpha ,\cos \alpha$ are the roots of the equation $c{x}^2+bx+a=0$, then $a,b,c$ satisfies

• A. $c^2+2ac+b^2=0$
• B. $b^2-2ac-c^2=0$
• C. $2ac-b^2-c^2=0$
• D. $b^2+2ac-c^2=0$

Answer 1

Answer: (B)

$b^2-2ac-c^2=0$

If $\sin \alpha ,\cos \alpha$ are the roots of the equation,
$c{x}^2+bx+a=0$

Since , $\sin \alpha ,\cos \alpha$ are the roots of the equation,

$\sin \alpha +\cos \alpha =\frac{-b}{c}$

$\sin \alpha \times \cos \alpha =\frac{a}{c}$

Now, on Squaring first equation, we have

${(\sin \alpha +\cos \alpha )}^2={\left(\frac{-b}{c}\right)}^2$

$({\sin }^2\alpha +{\cos }^2\alpha +2\sin \alpha \cos \alpha )=\frac{b^2}{c^2}$

$(1+2\sin \alpha \cos \alpha )=\frac{b^2}{c^2}$

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

On simplifying this , we get
$(1+\frac{ac}{c^2})=\frac{b^2}{c^2}$

$\left(\frac{c^2+ac}{c^2}\right)=\frac{b^2}{c^2}$

$b^2-2ac-c^2=0$