Complex Numbers And Quadratic Equations question 522
Question: If $ \alpha $ and $ \beta $ be the roots of the equation $ 2x^{2}+2(a+b)x+a^{2}+b^{2}=0 $ , then the equation whose roots are $ {{(\alpha +\beta )}^{2}} $ and $ {{(\alpha -\beta )}^{2}} $ is
Options:
A) $ x^{2}-2abx-{{(a^{2}-b^{2})}^{2}}=0 $
B) $ x^{2}-4abx-{{(a^{2}-b^{2})}^{2}}=0 $
C) $ x^{2}-4abx+{{(a^{2}-b^{2})}^{2}}=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Sum of roots $ \alpha +\beta =-(a+b) $ and $ \alpha \beta =\frac{a^{2}+b^{2}}{2} $
Þ $ {{(\alpha +\beta )}^{2}}={{(a+b)}^{2}} $ and $ {{(\alpha -\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta $ = $ 2ab-(a^{2}+b^{2})=-{{(a-b)}^{2}} $ Now the required equation whose roots are $ {{(\alpha +\beta )}^{2}} $ and $ {{(\alpha -\beta )}^{2}} $ $ x^{2}-{{{(\alpha +\beta )}^{2}}+{{(\alpha -\beta )}^{2}}}x+{{(\alpha +\beta )}^{2}}{{(\alpha -\beta )}^{2}}=0 $
$ \Rightarrow x^{2}-{{{(a+b)}^{2}}-{{(a-b)}^{2}}}x-{{(a+b)}^{2}}{{(a-b)}^{2}}=0 $
$ \Rightarrow x^{2}-4abx-{{(a^{2}-b^{2})}^{2}}=0 $
BETA
