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Q.
If $\alpha , \beta, \gamma $ are the roots of the equation $2x^3-3x^2+6x+1=0$ then $\alpha^2 +\beta^2+\gamma^2$ is equal to
KCETKCET 2005Complex Numbers and Quadratic Equations
Solution:
Since, $\alpha, \beta, \gamma$ are the roots of the equation
$2 x^{3}-3 x^{2}+6 x+1=0$, then
$\alpha+\beta+\gamma =+\frac{3}{2} \,\,\,\,\,\dots(i)$
$\alpha \beta+\beta \gamma+\gamma \alpha =3 \,\,\,\,\dots(ii)$
$\alpha \beta \gamma =-\frac{1}{2}\,\,\,\,\dots(iii)$
On squaring both sides Eq. (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 Eq. (ii) ]
$=\frac{9}{4}-6=-\frac{15}{4}$