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 :
We have,\[{{\alpha }^{2}}=5\alpha -3\] \[\Rightarrow \] \[{{\alpha }^{2}}-5\alpha +3=0\] \[\Rightarrow \] \[\alpha =\frac{5\pm \sqrt{13}}{2}\] Similarly,\[{{\beta }^{2}}=5\beta -3\] \[\Rightarrow \] \[\beta =\frac{5\pm \sqrt{13}}{2}\] Since, \[\alpha \ne \beta \] \[\therefore \] \[\alpha =\frac{5+\sqrt{13}}{2}\]and\[\beta =\frac{5-\sqrt{13}}{2}\] or \[\alpha =\frac{5-\sqrt{13}}{2}\] and \[\beta =\frac{5+\sqrt{13}}{2}\] \[\therefore \] \[{{\alpha }^{2}}+{{\beta }^{2}}=\frac{50+26}{4}=19\] and \[\alpha \beta =\frac{1}{4}(25-13)=3\] Thus, the equation having\[\frac{\alpha }{\beta }\]and\[\frac{\beta }{\alpha }\]as its roots is \[{{x}^{2}}-x\left( \frac{\alpha }{\beta }+\frac{\beta }{\alpha } \right)+\frac{\alpha \beta }{\alpha \beta }=0\] \[\Rightarrow \] \[{{x}^{2}}-x\left( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right)+1=0\] \[\Rightarrow \] \[3{{x}^{2}}-19x+3=0\]
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