Gujarati
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

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