A) \[3{{x}^{2}}-25x+3=0\]
B) \[{{x}^{2}}+5x-3=0\]
C) \[{{x}^{2}}-5x+3=0\]
D) \[3{{x}^{2}}-19x+3=0\]
Correct Answer: D
Solution :
\[{{\alpha }^{2}}-5\alpha +3=0\] ?..(i) \[{{\beta }^{2}}-5\beta +3=0\] ?..(ii) From (i) - (ii), Þ \[({{\alpha }^{2}}-{{\beta }^{2}})-5\alpha +5\beta =0\] Þ \[{{\alpha }^{2}}-{{\beta }^{2}}=5(\alpha -\beta )\]\[\Rightarrow \alpha +\beta =5\] From (i) + (ii), Þ \[({{\alpha }^{2}}+{{\beta }^{2}})-5(\alpha +\beta )+6=0\] Þ \[({{\alpha }^{2}}+{{\beta }^{2}})-5.5+6=0\] Þ \[{{\alpha }^{2}}+{{\beta }^{2}}=19\] Then \[{{(\alpha +\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \] \[\Rightarrow 25=19+2\alpha \beta \] \[\Rightarrow \alpha \beta =3\] then the equation, whose roots are \[\frac{\alpha }{\beta }\] and \[\frac{\beta }{\alpha }\], is \[{{x}^{2}}-x\left( \frac{\alpha }{\beta }+\frac{\beta }{\alpha } \right)+\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=0\] Þ \[{{x}^{2}}-x\left( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right)+1=0\] \[\Rightarrow {{x}^{2}}-x.\frac{19}{3}+1=0\]Þ \[3{{x}^{2}}-19x+3=0\].
You need to login to perform this action.
You will be redirected in
3 sec
