A) \[{{a}^{2}}+{{b}^{2}}=2ac\]
B) \[{{b}^{2}}-{{c}^{2}}=2ab\]
C) \[{{b}^{2}}-{{a}^{2}}=2ac\]
D) \[{{b}^{2}}+{{c}^{2}}=2ab\]
Correct Answer: C
Solution :
Let \[\sin \,\,\alpha \] and \[\cos \,\,\alpha \] be the roots of \[{{\operatorname{ax}}^{2}}+bx+c=0\] Now, \[\sin \alpha + cos \alpha = \frac{-b}{a}\] and \[\sin \alpha + cos \alpha = \frac{c}{a}\] Consider \[\sin \,\alpha + cos \alpha = \frac{-b}{a}\] Squaring both side, \[{{\left( \sin \alpha + cos \alpha \right)}^{2}} =\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \,\, {{\sin }^{2}}\alpha + co{{s}^{2}} \alpha + 2 sin \alpha cos \alpha = \frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \,\,1+\frac{2c}{a}=\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \,\,\,\frac{a+2c}{a}=\frac{{{b}^{2}}}{{{a}^{2}}}\Rightarrow a+2c=\frac{{{b}^{2}}}{a}\] \[\Rightarrow \,\,{{a}^{2}}+2\,\,ac={{b}^{2}}\Rightarrow {{b}^{2}}-{{a}^{2}}=2\,ac\]
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