Mathematics - Application of Derivative Question with Solution | TestHub

$\begin{array}{rl}& \alpha +{\alpha }^2=3\mathrm{a}\&\alpha \times {\alpha }^2=2{\mathrm{a}}^2\\ & \downarrow \\ & {\left(\alpha +{\alpha }^2\right)}^3=27{\mathrm{a}}^3\\ & \Rightarrow 2{\mathrm{a}}^2+4{\mathrm{a}}^4+3(3\mathrm{a})\left(2{\mathrm{a}}^2\right)=27{\mathrm{a}}^3\\ & \Rightarrow 2+4{\mathrm{a}}^2+18\mathrm{a}=27\mathrm{a}\\ & \Rightarrow 4{\mathrm{a}}^2-9\mathrm{a}+2=0\\ & \Rightarrow 4{\mathrm{a}}^2-8\mathrm{a}-\mathrm{a}+2=0\\ & \Rightarrow (4\mathrm{a}-1)(\mathrm{a}-2)=0\Rightarrow \mathrm{a}=2\\ & \text{ so }6{\mathrm{x}}^2-36\mathrm{x}+48=0\end{array}$ $\Rightarrow x^2-6x+8=0$ ...(1) If we take $\mathrm{a}=\frac{1}{4}$ then $\alpha =\frac{1}{2}$ which is not possible