"If the function $f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a} 0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$, then $\alpha$ and $\alpha^2$ are the roots of the equation :"

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Accepted Answer

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$$\begin{aligned} & f(x)=6 x^2-18 a x+12 a^2 \\ & =6\left(x^2-3 a+2 a^2\right) \\ & =6(x-a)(x-2 a)=0 \\ & x=a, 2 a \end{aligned}$$

$a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2$

$$\begin{array}{lll} a>0 & \therefore & \alpha=2 \\ & & \alpha^2=4 \end{array}$$

$\therefore x^2-6 x+8=0$ is the required quadratic equation.

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If the function $f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a}> 0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$, then $\alpha$ and $\alpha^2$ are the roots of the equation :

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If the function $f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a}> 0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$, then $\alpha$ and $\alpha^2$ are the roots of the equation :

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