Question

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 :

• A. $ - \frac{15}{4} $
• B. $ \frac{15}{4} $
• C. $ - \frac{9}{4} $
• D. $ 4$

Answer 1

Answer: (A)

$2 x^{3}-3 x^{2}+6 x+1=0$
sum of roots $=\alpha+\beta+\gamma=-\frac{-3}{2}=\frac{3}{2}$
product of roots taken two at a time
$=\alpha \beta+\beta \gamma+\gamma \alpha=\frac{6}{2}=3$
$\alpha^{2}+\beta^{2}+\gamma^{2} =(\alpha+\beta+\gamma)^{2}-2(\alpha \beta+\beta \gamma+\gamma \alpha)$
$=\left(\frac{3}{2}\right)^{2}-2(3)$
$=\frac{9}{4}-6$
$=-\frac{15}{4}$