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)

Since, $\alpha ,\beta ,\gamma$ are the roots of the equation
$2x^3-3x^2+6x+1=0$, 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) ]
$=\frac{9}{4}-6=-\frac{15}{4}$