If the set of all values of a , for which the equation 5 x^3-15 x-a=0 has three distinct real roots, is the interval (alpha, beta), then beta-2 alpha is equal to \_\_\_\_ .

\ 5 x^3-15 x-a=0 f(x)=5 x^3-15 x f^prime(x)=15 x^2-15=15(x-1)(x+1) a in (-10,10) alpha=-10, beta=10 beta-2 alpha=10+20=30\

If the set of all values of a , for which the equation ... | (Mathematics)

$\begin{aligned}& 5 x^3-15 x-a=0 \\& f(x)=5 x^3-15 x \\& f^{\prime}(x)=15 x^2-15=15(x-1)(x+1)\end{aligned}$ $\begin{aligned}& \mathrm{a} \in(-10,10) \\& \alpha=-10, \beta=10 \\& \beta-2 \alpha=10+20=30\end{aligned}$

Explanation

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