A) \[\beta \gamma \]
B) \[0\]
C) \[\alpha \gamma \]
D) \[\alpha \beta \]
Correct Answer: A
Solution :
\[\alpha {{x}^{2}}+2\beta x+\gamma =0\] Let\[\beta =\alpha t,\gamma =\alpha {{t}^{2}}\] \[\therefore \]\[\alpha {{x}^{2}}+2\alpha tx+\alpha {{t}^{2}}=0\] \[\Rightarrow {{x}^{2}}+2tx+{{t}^{2}}=0\] \[\Rightarrow {{(x+t)}^{2}}=0\] \[\Rightarrow x=-t\] it must be root of equation \[{{x}^{2}}+x1=0\] \[\therefore {{t}^{2}}-t-1=0\] (1) Now \[\alpha (\beta +\gamma )={{\alpha }^{2}}(t+{{t}^{2}})\] Option \[\beta \gamma =\alpha t.\alpha {{t}^{2}}={{\alpha }^{2}}{{t}^{3}}={{a}^{2}}({{t}^{2}}+t)\] (from equation 1)You need to login to perform this action.
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