A) \[{{a}^{2}}-{{b}^{2}}+2ac=0\]
B) \[{{(a+c)}^{2}}={{b}^{2}}-{{c}^{2}}\]
C) \[{{a}^{2}}+{{b}^{2}}-2ac=0\]
D) \[{{(a-c)}^{2}}={{b}^{2}}+{{c}^{2}}\]
Correct Answer: A
Solution :
Given equation is \[a{{x}^{2}}+bx+c=0\] \[\therefore \] \[\sin \alpha +\cos \alpha =-\frac{b}{a}\] and \[\sin \alpha \cos \alpha =\frac{c}{a}\] \[\Rightarrow \]\[{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +2\sin \alpha \cos \alpha =\frac{{{b}^{2}}}{{{a}^{2}}}\] and \[\sin \alpha \cos \alpha =\frac{c}{a}\] \[\Rightarrow \]\[1+\frac{2c}{a}=\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \]\[{{a}^{2}}+2ac={{b}^{2}}\] \[\Rightarrow \] \[{{a}^{2}}-{{b}^{2}}+2ac=0\]
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