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

<p>To find the sum of the fourth powers of the roots $\alpha_\theta$ and $\beta_\theta$ of the quadratic equation $2x^2 + (\cos \theta)x - 1 = 0$, we start analyzing the expression $\alpha_\theta^4 + \beta_\theta^4$.</p> <p><p>The equation can be rewritten with its roots using:</p> <p>$ \alpha + \beta = -\frac{\cos \theta}{2}, \quad \alpha \beta = -\frac{1}{2} $</p></p> <p><p>We need to calculate $\alpha^2 + \beta^2$ and $\alpha^2 \beta^2$:</p> <p>$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{\cos \theta}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{\cos^2 \theta}{4} + 1 $</p> <p>$ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} $</p></p> <p><p>Substitute these into:</p> <p>$ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - \frac{1}{2} $</p></p> <p><p>Maximize and minimize $\left(\frac{\cos^2 \theta}{4} + 1\right)^2$:</p></p> <p><p>Zero of $\cos\theta$ leads to:</p> <p>$ \left(\frac{0^2}{4} + 1\right)^2 = 1 $</p></p> <p><p>Max value $\cos^2 \theta = 1$:</p> <p>$ \left(\frac{1}{4} + 1\right)^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16} $</p></p> <p><p>Substitute back:</p> <p>$ \text{Max: } \frac{25}{16} - \frac{1}{2} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16} $</p> <p>$ \text{Min: } 1 - \frac{1}{2} = \frac{1}{2} $</p></p> <p><p>Finally, compute $16(M + m)$:</p> <p>$ 16\left(\frac{17}{16} + \frac{1}{2}\right) = 16\left(\frac{17}{16} + \frac{8}{16}\right) = 16 \times \frac{25}{16} = 25 $</p></p>