A) \[\frac{10}{3}\]
B) \[\frac{3}{10}\]
C) \[\frac{2}{3}\]
D) None of these
Correct Answer: A
Solution :
[a] \[{{x}^{2}}+2x+3=0\] \[\therefore \alpha +\beta =-2\And \alpha .\beta =3\] Now \[\frac{{{\alpha }^{2}}}{\beta }+\frac{{{\beta }^{2}}}{\alpha }=\frac{{{\alpha }^{3}}+{{\beta }^{2}}}{\alpha .\beta }=\frac{{{(\alpha +\beta )}^{3}}-3\alpha \beta (\alpha +\beta )}{\alpha .\beta }\] \[=\frac{{{(-2)}^{3}}-3.3(-2)}{3}=\frac{-8+18}{3}=\frac{10}{3}\] Hence, option [a] is correct.
You need to login to perform this action.
You will be redirected in
3 sec
