Exact set of values of a for which {x^3} | vedclass

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Question

Equation becomes $4 a^{2}+6 a x-\left(x^{4}+x^{3}-2 x^{2}\right)=0$

$a=\frac{-6 x \pm \sqrt{36 x^{2}+16\left(x^{4}+x^{3}-2 x^{2}\right)}}{8}$

$a

Vedclass · Accepted Answer

$\left[ { - \frac{1}{8},\frac{1}{2}} \right]$

Vedclass · Answer

Equation becomes $4 a^{2}+6 a x-\left(x^{4}+x^{3}-2 x^{2}\right)=0$

$a=\frac{-6 x \pm \sqrt{36 x^{2}+16\left(x^{4}+x^{3}-2 x^{2}\right)}}{8}$

$a=\frac{-x^{2}}{2}-x$ and $a=\frac{x^{2}}{2}-\frac{x}{2}$

$\mathrm{x}^{2}+2 \mathrm{x}+2 \mathrm{a}=0$ or $\mathrm{x}^{2}-\mathrm{x}-2 \mathrm{a}=0$

$\mathrm{D}_{1} \geq 0$  and $ \mathrm{D}_{2} \geq 0$

$\mathrm{a} \in\left[-\frac{1}{8}, \frac{1}{2}\right]$

Exact set of values of $a$ for which ${x^3}(x + 1) = 2(x + a)(x + 2a)$ is having four real solutions is

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