Complex Numbers And Quadratic Equations question 620
Question: If $ \alpha ,\beta $ are the roots of $ ax^{2}+bx+c=0 $ and $ \alpha +\beta , $ $ {{\alpha }^{2}}+{{\beta }^{2}}, $ $ {{\alpha }^{3}}+{{\beta }^{3}} $ are in G.P., where $ \Delta =b^{2}-4ac $ , then [IIT Screening 2005]
Options:
A) $ \Delta \ne 0 $
B) $ b\Delta =0 $
C) $ cb\ne 0 $
D) $ c\Delta =0 $
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Answer:
Correct Answer: D
Solution:
$ {{({{\alpha }^{2}}+{{\beta }^{2}})}^{2}}=(\alpha +\beta )({{\alpha }^{3}}+{{\beta }^{3}}) $ $ {{( \frac{b^{2}-2ac}{a^{2}} )}^{2}}=( \frac{-b}{a} )( \frac{-b^{2}+3abc}{a^{3}} ) $
Þ $ 4a^{2}c^{2}=acb^{2} $
Þ $ ac(b^{2}-4ac)=0 $ As $ a\ne 0\Rightarrow c\Delta =0 $
BETA
