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
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 \[\frac{{{\alpha }^{2}}}{b{a}'-{b}'c}=\frac{\alpha }{c{c}'-a{a}'}=\frac{1}{a{b}'-b{c}'}\] \[{{(cc'-aa')}^{2}}=(ba'-cb')(ab'-bc')\].
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