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4-2.Quadratic Equations and Inequations
hard
यदि $\alpha, \beta $ $\gamma$ समीकरण $2{x^3} - 3{x^2} + 6x + 1 = 0$ के मूल हों, तो ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$ का मान है
A
-$\frac{{15}}{4}$
B
$\frac{{15}}{4}$
C
$\frac{9}{4}$
D
$4$
Solution
(a) दिया गया समीकरण $2{x^3} – 3{x^2} + 6x + 1 = 0$है।
$\alpha + \beta + \gamma = \frac{3}{2}$, $\alpha \beta \gamma = \frac{{ – 1}}{2}$, $\Sigma \alpha \beta = 3$
$({\alpha ^2} + {\beta ^2} + {\gamma ^2}) = {(\alpha + \beta + \gamma )^2} – 2(\Sigma \alpha \beta )$
= ${\left( {\frac{3}{2}} \right)^2} – 2.3$= $\frac{9}{4} – 6 = \frac{{ – 15}}{4}$.
Standard 11
Mathematics
