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)

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