A) \[-\frac{15}{4}\]
B) \[\frac{15}{4}\]
C) \[\frac{9}{4}\]
D) 4
Correct Answer: A
Solution :
Since \[\alpha ,\,\beta ,\,\gamma \], are the roots of the equation\[2{{x}^{3}}-3{{x}^{2}}+6x+1=0\], then \[\alpha +\beta +\gamma =+\frac{3}{2}\] ... (i) \[\alpha \beta +\beta \gamma +\gamma \alpha =3\] ... (ii) \[\alpha \,\beta \,\,\gamma =-\frac{1}{2}\] ?. (iii) On squaring equation (i), we get \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2\,(\alpha \,\beta +\beta \,\gamma +\gamma \alpha )=\frac{9}{4}\] \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}=\frac{9}{4}-2\,(3)\] [from (ii)] \[=\frac{9}{4}-6=-\frac{15}{4}\]
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