A) \[b{{x}^{2}}+ax+1=0\]
B) \[b{{x}^{2}}-ax+1=0\]
C) \[b{{x}^{2}}+(a+2b)x+a+b+1=0\]
D) \[b{{x}^{2}}-(a+2b)x+a+b+1=0\]
Correct Answer: D
Solution :
Let\[\alpha '=\alpha /(\alpha -1),\beta '=\beta /(\beta -1)\] \[\Rightarrow \]\[\alpha =\alpha '/(\alpha '-1),\beta =\beta '/(\beta '-1)\] \[\Rightarrow \]\[1/\alpha =(\alpha '-1)/\alpha ,1/\beta =(\beta '-1)/\beta '\] Then equation whose roofs are \[1/\alpha ,1/\beta ,\]is \[{{x}^{2}}-(1/\alpha +1/\beta )x+1/\alpha \beta =0\] \[\Rightarrow \]\[{{x}^{2}}-[(\alpha '-1)/\alpha '+(\beta '-1)/\beta ']x+\] \[[(\alpha '-1)(\beta '-1)/\alpha '\beta ']=0\] \[\Rightarrow \]\[[\alpha 'b'{{x}^{2}}-[2\alpha 'b'-(\alpha '+b')]\] \[x+\alpha 'b'-(\alpha '+b')+1=0\] \[\Rightarrow \]\[b{{x}^{2}}-(a+2b)x+a+b+1=0\]
You need to login to perform this action.
You will be redirected in
3 sec
