Complex Numbers And Quadratic Equations question 618
Question: If one root of the equation $ x^{2}+px+q=0 $ is the square of the other, then [IIT Screening 2004]
Options:
A) $ p^{3}+q^{2}-q(3p+1)=0 $
B) $ p^{3}+q^{2}+q(1+3p)=0 $
C) $ p^{3}+q^{2}+q(3p-1)=0 $
D) $ p^{3}+q^{2}+q(1-3p)=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
Let root of the given equation $ x^{2}+px+q=0 $ are $ \alpha $ and $ {{\alpha }^{2}} $ . Now, $ \alpha .{{\alpha }^{2}}={{\alpha }^{3}}=q, $ $ \alpha +{{\alpha }^{2}}=-p $ Cubing both sides, $ {{\alpha }^{3}}+{{({{\alpha }^{2}})}^{3}}+3\alpha .{{\alpha }^{2}}(\alpha +{{\alpha }^{2}})=-p^{3} $ $ q+q^{2}+3q(-p)=-p^{3} $ $ p^{3}+q^{2}+q(1-3p)=0 $ .
BETA
