Complex Numbers And Quadratic Equations question 638
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( \frac{1}{\sqrt{2}}<a<1 ) $ and $ {{\beta }^{2}}\ne 1, $ then $ {{\beta }^{2}}= $
Options:
A) $ a^{2} $
B) $ \frac{1-a^{2}}{a^{2}} $
C) $ 1-a^{2} $
D) $ 1+a^{2} $
Show Answer
Answer:
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}} $
BETA
