- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
hard
If $\alpha, \beta $ and $\gamma$ are the roots of the equation $2{x^3} - 3{x^2} + 6x + 1 = 0$, then ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$ is equal to
A
-$\frac{{15}}{4}$
B
$\frac{{15}}{4}$
C
$\frac{9}{4}$
D
$4$
Solution
(a) Given equation $2{x^3} – 3{x^2} + 6x + 1 = 0$
$\alpha + \beta + \gamma = \frac{3}{2}$, $\alpha \beta \gamma = \frac{{ – 1}}{2}$,
$\Sigma \alpha \beta = 3$
$({\alpha ^2} + {\beta ^2} + {\gamma ^2}) = {(\alpha + \beta + \gamma )^2} – 2(\Sigma \alpha \beta )$
= ${\left( {\frac{3}{2}} \right)^2} – 2.3$= $ \frac{9}{4} – 6 = \frac{{ – 15}}{4}$.
Standard 11
Mathematics
