Complex Numbers And Quadratic Equations question 535
Question: If $ \alpha ,\beta $ be the roots of $ x^{2}-px+q=0 $ and $ {\alpha }’,{\beta }’ $ be the roots of $ x^{2}-{p}‘x+{q}’=0 $ , then the value of $ {{(\alpha -\alpha ‘)}^{2}}+{{(\beta -{\alpha }’)}^{2}}+{{(a-{\beta }’)}^{2}}+{{(\beta -{\beta }’)}^{2}} $ is
Options:
A) $ 2{p^{2}-2q+{{{p}’}^{2}}-2{q}’-p{p}’} $
B) $ 2{p^{2}-2q+{{{p}’}^{2}}-2{q}’-q{q}’} $
C) $ 2{p^{2}-2q-{{{p}’}^{2}}-2{q}’-p{p}’} $
D) $ 2{p^{2}-2q-{{{p}’}^{2}}-2{q}’-q{q}’} $
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Answer:
Correct Answer: A
Solution:
As given, $ \alpha +\beta =p, $ $ \alpha \beta =q,{\alpha }’+{\beta }’={p}’,{\alpha }’{\beta }’=q’ $ Now, $ {{(\alpha -\alpha ‘)}^{2}}+{{(\beta -\alpha ‘)}^{2}}+{{(\alpha -{\beta }’)}^{2}}+{{(\beta -{\beta }’)}^{2}} $ $ =2({{\alpha }^{2}}+{{\beta }^{2}})+2(\alpha {{’}^{2}}+\beta {{’}^{2}})-2\alpha ‘(\alpha +\beta )-2\beta ‘(a+\beta ) $ $ =2{ {{(\alpha +\beta )}^{2}}-2\alpha \beta +{{({\alpha }’+{\beta }’)}^{2}}-2\alpha ‘\beta ‘-(\alpha +\beta )({a}’+\beta ‘) } $ $ =2{ p^{2}-2q+{{{{p}’}}^{2}}-2{q}’-p{p}’ } $ .
BETA
