• # question_answer If $\sin \alpha ,\cos \alpha$ are the roots of the equation $a{{x}^{2}}+bx+c=0$, then [MP PET 1993] A) ${{a}^{2}}-{{b}^{2}}+2ac=0$ B) ${{(a-c)}^{2}}={{b}^{2}}+{{c}^{2}}$ C) ${{a}^{2}}+{{b}^{2}}-2ac=0$ D) ${{a}^{2}}+{{b}^{2}}+2ac=0$

Correct Answer: A

Solution :

As given, $\sin \alpha +\cos \alpha =-\frac{b}{a},$$\sin \alpha \cos \alpha =\frac{c}{a}$ To eliminate$\alpha$, we have $1={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha ={{(\sin \alpha +\cos \alpha )}^{2}}-2\sin \alpha \cos \alpha$     $=\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{a}\,\,\Rightarrow {{a}^{2}}-{{b}^{2}}+2ac=0$

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