A) \[{{x}^{2}}+2x+1=0\]
B) \[9{{x}^{2}}+2x+1=0\]
C) \[9{{x}^{2}}-2x+1=0\]
D) \[9{{x}^{2}}+2x-1=0\]
Correct Answer: B
Solution :
\[\alpha ,\beta \] be the roots of \[{{x}^{2}}-2x+3=0\], then \[\alpha +\beta =2\] and \[\alpha \beta =3\]. Now required equation whose roots are \[\frac{1}{{{\alpha }^{2}}},\frac{1}{{{\beta }^{2}}}\]is \[{{x}^{2}}-\left( \frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}} \right)x+\frac{1}{{{\alpha }^{2}}{{\beta }^{2}}}=0\] Þ \[{{x}^{2}}-\left( -\frac{2}{9} \right)x+\frac{1}{9}=0\]Þ \[9{{x}^{2}}+2x+1=0\].
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