A) \[3{{x}^{2}}+19x+3=0\]
B) \[3{{x}^{2}}-19x+3=0\]
C) \[3{{x}^{2}}-19x-3=0\]
D) \[{{x}^{2}}-16x+1=0\]
Correct Answer: B
Solution :
The equation having \[\alpha \] and \[\beta \] as its roots, is \[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0\]. Since, \[{{\alpha }^{2}}=5\alpha -3\Rightarrow {{\alpha }^{2}}-5\alpha +3=0\] And \[{{\beta }^{2}}=5\beta -3\Rightarrow {{\beta }^{2}}-5\beta +3=0\] These two equations show that \[\alpha \] and \[\beta \] are the roots of the equation \[{{x}^{2}}-5x+3=0\] \[\therefore \alpha +\beta =5\] and \[\alpha \beta =3\] Now, \[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta }\] \[=\frac{25-3}{3}=\frac{19}{3}\] and \[\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=1\] Thus, the equation having \[\frac{\alpha }{\beta }\] and \[\frac{\beta }{\alpha }\] as its roots, is given by \[{{x}^{2}}-\left( \frac{\alpha }{\beta }-\frac{\beta }{\alpha } \right)x+\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=0\] \[\Rightarrow \] \[{{x}^{2}}-\frac{19}{3}x+1=0\] \[\Rightarrow \] \[3{{x}^{2}}-19x+3=0\]
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