A) \[{{a}^{2}}\]
B) \[\frac{1-{{a}^{2}}}{{{a}^{2}}}\]
C) \[1-{{a}^{2}}\]
D) \[1+{{a}^{2}}\]
Correct Answer: B
Solution :
\[{{\alpha }^{2}}+{{\beta }^{2}}=\frac{1}{{{a}^{2}}}\] and \[{{\alpha }^{2}}{{\beta }^{2}}=\frac{1-{{a}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \,\,{{\alpha }^{2}}+{{\beta }^{2}}-1={{\alpha }^{2}}{{\beta }^{2}}\Rightarrow ({{\alpha }^{2}}-1)({{\beta }^{2}}-1)=0\] \[\because \,\,\,\,{{\beta }^{2}}\ne 1\Rightarrow {{\alpha }^{2}}=1,\] so, \[{{\beta }^{2}}=\frac{1-{{a}^{2}}}{{{a}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
