A) \[{{n}^{2}}+2\]
B) \[-{{n}^{2}}\]
C) \[{{n}^{2}}\]
D) \[2{{n}^{2}}\]
Correct Answer: C
Solution :
Equation \[{{x}^{2}}-{{(1+n)}^{2}}x+\frac{1}{2}(1+{{n}^{2}}+{{n}^{4}})=0\] has roots\[\alpha \]and\[\beta \]. \[\therefore \] \[\alpha +\beta =(1+{{n}^{2}})\] and \[\alpha \beta =\frac{1+{{n}^{2}}+{{n}^{4}}}{2}\] Now, \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[={{(1+{{n}^{2}})}^{2}}-2\left( \frac{1+{{n}^{2}}+{{n}^{4}}}{2} \right)\] \[=1+{{n}^{4}}+2{{n}^{2}}-1-{{n}^{2}}-{{n}^{4}}\] \[={{n}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
