A) \[\frac{{{B}^{2}}-2AC}{{{A}^{2}}}\]
B) \[\frac{2AC-{{B}^{2}}}{{{A}^{2}}}\]
C) \[\frac{{{B}^{2}}-4AC}{{{A}^{2}}}\]
D) None of these
Correct Answer: B
Solution :
\[\alpha ,\beta \] are the roots of \[A{{x}^{2}}+Bx+C=0\]. So, \[\alpha +\beta =-\frac{B}{A}\]and \[\alpha \beta =\frac{C}{A}\] Again \[{{\alpha }^{2}},{{\beta }^{2}}\] are the roots of \[{{x}^{2}}+px+q=0\] then \[{{\alpha }^{2}}+{{\beta }^{2}}=-p\] and\[{{(\alpha \beta )}^{2}}=q\] Now \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] Þ \[{{\alpha }^{2}}+{{\beta }^{2}}={{\left( -\frac{B}{A} \right)}^{2}}-2\frac{C}{A}\] Þ \[-p=\frac{{{B}^{2}}-2AC}{{{A}^{2}}}\Rightarrow p=\frac{2AC-{{B}^{2}}}{{{A}^{2}}}\]
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