A) \[\frac{215}{27}\]
B) \[\frac{357}{21}\]
C) \[\frac{115}{28}\]
D) \[\frac{325}{31}\]
Correct Answer: A
Solution :
| \[\alpha\] and \[\beta\]are the zeroes of quadratic polynomial. |
| \[f\left( x \right)=3{{x}^{2}}-5x-2\] |
| \[\therefore\] We know that, |
| \[\alpha +\beta =-\frac{b}{a}=\frac{-\left( -5 \right)}{3}=\frac{5}{3}\] |
| \[\alpha \,.\,\beta =\frac{c}{a}=-\frac{2}{3}\] |
| \[{{\alpha }^{3}}+{{\beta }^{3}}=\left( \alpha +\beta \right)\left( {{\alpha }^{2}}-\alpha \,.\,\beta +{{\beta }^{2}} \right)\] |
| \[=\left( \alpha +\beta \right)\left[ {{\left( \alpha +\beta \right)}^{2}}-3\alpha \,.\,\beta \right]\] |
| \[=\frac{5}{3}\left[ {{\left( \frac{5}{3} \right)}^{2}}-3\times \left( -\frac{2}{3} \right) \right]\] |
| \[=\frac{125}{27}+2\times \frac{5}{3}\] |
| \[=\frac{125+90}{27}=\frac{215}{27}\] |
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