The equation 16x^4 + 16x^3 - 4x - 1 = 0 has a | vedclass

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(D) Given the equation $f(x) = 16x^4 + 16x^3 - 4x - 1 = 0$. Since it has a multiple root,let $\alpha$ be the multiple root. Then $f(\alpha) = 0$ and $

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$64$

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(D) Given the equation $f(x) = 16x^4 + 16x^3 - 4x - 1 = 0$. Since it has a multiple root,let $\alpha$ be the multiple root. Then $f(\alpha) = 0$ and $f'(\alpha) = 0$. $f'(x) = 64x^3 + 48x^2 - 4$. Setting $f'(\alpha) = 0$ gives $16\alpha^3 + 12\alpha^2 - 1 = 0$. From $f(\alpha) = 16\alpha^4 + 16\alpha^3 - 4\alpha - 1 = 0$,we substitute $16\alpha^3 = 1 - 12\alpha^2$ to get $16\alpha^4 + (1 - 12\alpha^2) - 4\alpha - 1 = 0$,which simplifies to $16\alpha^4 - 12\alpha^2 - 4\alpha = 0$. Factoring gives $4\alpha(4\alpha^3 - 3\alpha - 1) = 0$. Since $\alpha 
eq 0$,we solve $4\alpha^3 - 3\alpha - 1 = 0$,which factors as $(\alpha - 1)(2\alpha + 1)^2 = 0$. Testing $\alpha = 1$ in $f(x)$ gives $16+16-4-1 
eq 0$. Thus,$\alpha = -\frac{1}{2}$ is the multiple root. Dividing $f(x)$ by $(x + \frac{1}{2})^2 = (x^2 + x + \frac{1}{4})$,we get $16x^2 - 4 = 0$,so $x = \pm \frac{1}{2}$. The roots are $\alpha = -\frac{1}{2}, \beta = -\frac{1}{2}, \gamma = -\frac{1}{2}, \delta = \frac{1}{2}$. Then $\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = (-2)^4 + (-2)^4 + (-2)^4 + (2)^4 = 16 + 16 + 16 + 16 = 64$.

The equation $16x^4 + 16x^3 - 4x - 1 = 0$ has a multiple root. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation,then $\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} =$

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