Complex Numbers And Quadratic Equations question 589
Question: If $ \alpha ,\beta $ are the roots of the equation $ x^{2}-(1+n^{2})x+\frac{1}{2}(1+n^{2}+n^{4})=0 $ then the value of $ {{\alpha }^{2}}+{{\beta }^{2}} $ is [RPET 1996]
Options:
A) $ 2n $
B) $ n^{3} $
C) $ n^{2} $
D) $ 2n^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \alpha +\beta =1+n^{2} $ ; $ \alpha \beta =\frac{1}{2}(1+n^{2}+n^{4}) $ \ $ {{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $ $ ={{(1+n^{2})}^{2}}-2.\frac{1}{2}(1+n^{2}+n^{4}) $ $ =1+n^{4}+2n^{2}-1-n^{2}-n^{4} $ Þ $ {{\alpha }^{2}}+{{\beta }^{2}}=n^{2} $ .
BETA
