Question
If $$\alpha ,\beta $$ are real and $${\alpha ^2},{\beta ^2}$$ are the roots of the equation $${a^2}{x^2} - x + 1 - {a^2} = 0\left( {\frac{1}{{\sqrt 2 }} < a < 1} \right)$$ and $${\beta ^2} \ne 1,{\text{then }}{\beta ^2} = $$
A.
$$a^2$$
B.
$$\frac{{1 - {a^2}}}{{{a^2}}}$$
C.
$${1 - {a^2}}$$
D.
$${1 + {a^2}}$$
Answer :
$$\frac{{1 - {a^2}}}{{{a^2}}}$$
Solution :
$$\eqalign{
& {\alpha ^2} + {\beta ^2} = \frac{1}{{{a^2}}}\,\,{\text{and }}{\alpha ^2}{\beta ^2} = \frac{{1 - {a^2}}}{{{a^2}}} \cr
& \Rightarrow {\alpha ^2} + {\beta ^2} - 1 = {\alpha ^2}{\beta ^2} \cr
& \Rightarrow \left( {{\alpha ^2} - 1} \right)\left( {{\beta ^2} - 1} \right) = 0 \cr
& \because {\beta ^2} \ne 1, \cr
& \Rightarrow {\alpha ^2} = 1,\,\,{\text{so,}}\,\,{\beta ^2} = \frac{{1 - {a^2}}}{{{a^2}}} \cr} $$