Let alpha_theta and beta_theta be the distinc | vedclass

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(B) Given the quadratic equation $2x^2 + (\cos 	heta)x - 1 = 0$. Sum of roots $\alpha_	heta + \beta_	heta = -\frac{\cos 	heta}{2}$ and product of

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$25$

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(B) Given the quadratic equation $2x^2 + (\cos 	heta)x - 1 = 0$. Sum of roots $\alpha_	heta + \beta_	heta = -\frac{\cos 	heta}{2}$ and product of roots $\alpha_	heta \beta_	heta = -\frac{1}{2}$. We need to find $\alpha_	heta^4 + \beta_	heta^4 = (\alpha_	heta^2 + \beta_	heta^2)^2 - 2(\alpha_	heta \beta_	heta)^2$. Since $\alpha_	heta^2 + \beta_	heta^2 = (\alpha_	heta + \beta_	heta)^2 - 2\alpha_	heta \beta_	heta = \frac{\cos^2 	heta}{4} + 1$. Thus,$\alpha_	heta^4 + \beta_	heta^4 = (\frac{\cos^2 	heta}{4} + 1)^2 - 2(-\frac{1}{2})^2 = (\frac{\cos^2 	heta}{4} + 1)^2 - \frac{1}{2}$. Let $u = \cos^2 	heta$,where $u \in [0, 1]$. Let $f(u) = (\frac{u}{4} + 1)^2 - \frac{1}{2}$. For $u = 0$,$f(0) = (1)^2 - \frac{1}{2} = \frac{1}{2} = m$. For $u = 1$,$f(1) = (\frac{1}{4} + 1)^2 - \frac{1}{2} = (\frac{5}{4})^2 - \frac{1}{2} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16} = M$. Therefore,$16(M + m) = 16(\frac{17}{16} + \frac{1}{2}) = 16(\frac{17 + 8}{16}) = 25$.

Let $\alpha_\theta$ and $\beta_\theta$ be the distinct roots of $2x^2 + (\cos \theta)x - 1 = 0$,where $\theta \in (0, 2\pi)$. If $m$ and $M$ are the minimum and the maximum values of $\alpha_\theta^4 + \beta_\theta^4$,then $16(M + m)$ equals:

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