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(A) Given the cubic equation $2x^3 - 3x^2 + 6x + 1 = 0$. By Vieta's formulas,for a cubic equation $ax^3 + bx^2 + cx + d = 0$,the sum of roots $\alpha

Vedclass · Accepted Answer

$-\frac{15}{4}$

Vedclass · Answer

(A) Given the cubic equation $2x^3 - 3x^2 + 6x + 1 = 0$. By Vieta's formulas,for a cubic equation $ax^3 + bx^2 + cx + d = 0$,the sum of roots $\alpha + \beta + \gamma = -\frac{b}{a}$ and the sum of roots taken two at a time $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$. Here,$a = 2, b = -3, c = 6, d = 1$. So,$\alpha + \beta + \gamma = -(\frac{-3}{2}) = \frac{3}{2}$ and $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{6}{2} = 3$. We know that $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$. Substituting the values: $\alpha^2 + \beta^2 + \gamma^2 = (\frac{3}{2})^2 - 2(3) = \frac{9}{4} - 6 = \frac{9 - 24}{4} = -\frac{15}{4}$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $2x^3 - 3x^2 + 6x + 1 = 0$,then $\alpha^2 + \beta^2 + \gamma^2$ is equal to

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