The exact set of values of a for which the eq | vedclass

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(D) The given equation is $x^4 + x^3 = 2(x^2 + 3ax + 2a^2)$.Rearranging the terms,we get $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = 0$.This can be factored as $

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

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(D) The given equation is $x^4 + x^3 = 2(x^2 + 3ax + 2a^2)$.Rearranging the terms,we get $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = 0$.This can be factored as $(x^2 + 2x + 2a)(x^2 - x - 2a) = 0$.For the equation to have four real solutions,both quadratic factors must have real roots.For $x^2 + 2x + 2a = 0$,the discriminant $D_1 = 2^2 - 4(1)(2a) = 4 - 8a \geq 0$,which implies $a \leq \frac{1}{2}$.For $x^2 - x - 2a = 0$,the discriminant $D_2 = (-1)^2 - 4(1)(-2a) = 1 + 8a \geq 0$,which implies $a \geq -\frac{1}{8}$.Combining these conditions,the set of values for $a$ is $[-\frac{1}{8}, \frac{1}{2}]$.

The exact set of values of $a$ for which the equation ${x^3}(x + 1) = 2(x + a)(x + 2a)$ has four real solutions is:

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