Complex Numbers And Quadratic Equations question 521
Question: If a root of the equation $ ax^{2}+bx+c=0 $ be reciprocal of a root of the equation then $ {a}‘x^{2}+{b}‘x+{c}’=0 $ , then [IIT 1968]
Options:
A) $ {{(c{c}’-a{a}’)}^{2}}=(b{a}’-c{b}’)(a{b}’-b{c}’) $
B) $ {{(b{b}’-a{a}’)}^{2}}=(c{a}’-b{c}’)(a{b}’-b{c}’) $
C) $ {{(c{c}’-a{a}’)}^{2}}=(b{a}’+c{b}’)(a{b}’+b{c}’) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ \alpha $ be a root of first equation, then $ \frac{1}{\alpha } $ be a root of second equation. Therefore $ a{{\alpha }^{2}}+b\alpha +c=0 $ and $ a’\frac{1}{{{\alpha }^{2}}}+{b}’\frac{1}{\alpha }+{c}’=0 $ or $ {c}’{{\alpha }^{2}}+{b}’\alpha +{a}’=0 $ Hence $ \frac{{{\alpha }^{2}}}{b{a}’-{b}‘c}=\frac{\alpha }{c{c}’-a{a}’}=\frac{1}{a{b}’-b{c}’} $ $ {{(cc’-aa’)}^{2}}=(ba’-cb’)(ab’-bc’) $ .
BETA
