A) \[{{b}^{2}}-{{a}^{2}}=2ac\]
B) \[{{a}^{2}}-{{b}^{2}}=2ac\]
C) \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]
D) \[{{a}^{2}}+{{b}^{2}}=2ac\]
Correct Answer: A
Solution :
| [a] \[\sin \theta \] and \[\cos \theta \] are the roots, |
| \[\sin \theta +\cos \theta =-\left( \frac{-b}{a} \right)\] |
| and \[\sin .\cos =\frac{c}{a}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{(\sin \theta +\cos \theta )}^{2}}={{\left( \frac{b}{a} \right)}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cdot \cos \theta =\frac{{{b}^{2}}}{{{a}^{2}}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,1+2\frac{c}{a}=\frac{{{b}^{2}}}{{{a}^{2}}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{a}^{2}}+2ac={{b}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{b}^{2}}-{{a}^{2}}=2ac\] |
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