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(C) Let $f(x) = 4x^4 + 4x^3 - 23x^2 - 12x + 36$. If $\alpha$ is a multiple root,then $f'(\alpha) = 0$. $f'(x) = 16x^3 + 12x^2 - 46x - 12$. Setting $f'

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$5$

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(C) Let $f(x) = 4x^4 + 4x^3 - 23x^2 - 12x + 36$. If $\alpha$ is a multiple root,then $f'(\alpha) = 0$. $f'(x) = 16x^3 + 12x^2 - 46x - 12$. Setting $f'(x) = 0$: $8x^3 + 6x^2 - 23x - 6 = 0$. Testing integer roots,for $x = -2$: $8(-8) + 6(4) - 23(-2) - 6 = -64 + 24 + 46 - 6 = 0$. So,$x = -2$ is a root. For $x = 1.5$ or $3/2$: $8(27/8) + 6(9/4) - 23(3/2) - 6 = 27 + 13.5 - 34.5 - 6 = 0$. So,$x = 1.5$ is a root. Checking $f(-2) = 4(16) + 4(-8) - 23(4) - 12(-2) + 36 = 64 - 32 - 92 + 24 + 36 = 0$. Checking $f(1.5) = 4(5.0625) + 4(3.375) - 23(2.25) - 12(1.5) + 36 = 20.25 + 13.5 - 51.75 - 18 + 36 = 0$. Since $\alpha > \beta$,$\alpha = 1.5$ and $\beta = -2$. Then $2\alpha - \beta = 2(1.5) - (-2) = 3 + 2 = 5$.

If $\alpha$ and $\beta$ $(\alpha > \beta)$ are the multiple roots of the equation $4x^4 + 4x^3 - 23x^2 - 12x + 36 = 0$,then $2\alpha - \beta = $

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