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4-2.Quadratic Equations and Inequations
normal
Exact set of values of $a$ for which ${x^3}(x + 1) = 2(x + a)(x + 2a)$ is having four real solutions is
A
$[-1,2]$
B
$[-3,7]$
C
$[-2,4]$
D
$\left[ { - \frac{1}{8},\frac{1}{2}} \right]$
Solution
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$
$\mathrm{a} \in\left[-\frac{1}{8}, \frac{1}{2}\right]$
Standard 11
Mathematics
