Complex Numbers And Quadratic Equations question 534
Question: If $ \alpha ,\beta $ are the roots of $ x^{2}+px+1=0 $ and $ \gamma ,\delta $ are the roots of $ x^{2}+qx+1=0 $ ,then $ q^{2}-p^{2} $ = [IIT 1978; DCE 2000]
Options:
A) $ (\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta ) $
B) $ (\alpha +\gamma )(\beta +\gamma )(\alpha -\delta )(\beta +\delta ) $
C) $ (\alpha +\gamma )(\beta +\gamma )(\alpha +\delta )(\beta +\delta ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
As given, $ \alpha +\beta =-p,\alpha \beta =1,\gamma +\delta =-q $ and $ \gamma \delta =1 $ Now, $ (\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta ) $ = $ {\alpha \beta -\gamma (\alpha +\beta )+{{\gamma }^{2}}}{\alpha \beta +\delta (\alpha +\beta )+{{\delta }^{2}}} $ = $ (1+p\gamma +{{\gamma }^{2}})(1-p\delta +{{\delta }^{2}}) $ $ =(p\gamma -q\gamma )(-p\delta -q\delta ) $ (Since $ \gamma $ is a root of $ x^{2}+qx+1=0 $ ) Þ $ {{\gamma }^{2}}+q\gamma +1=0\Rightarrow {{\gamma }^{2}}+1=-q\gamma $ and similarly $ {{\delta }^{2}}+1=-q\delta $ $ =-\gamma \delta (p-q)(p+q)=q^{2}-p^{2} $ .
BETA
