2017/18 AHSMC Part 1 Problem 14

The given equation can be written as $$a^2 - (2x^2 + 1)a - x + x^4 = 0$$ which has the solutions $a = x^2 - x$ and $a = x^2 + x + 1$. The solutions of the equation $x^2 - x - a = 0$ are different real numbers if $1 + 4a > 0$ or equivalently $a > -\dfrac{1}{4}$. The equation $x^2 + x + 1 - a = 0$ has different real roots if $1 - 4(1 - a) > 0$ or equivalently $a > \dfrac{3}{4}$. The above two equations cannot have common roots. Indeed, if for $x_0 \in \mathbb{R}$, $x_0^2 - x_0 - a = 0$ and $x_0^2 + x_0 + 1 - a = 0$ then $x_0 = -\dfrac{1}{2}$ and $a = \dfrac{3}{4}$, which is not convenient. Therefore, the solutions of the given equation are all real and different if $a > \dfrac{3}{4}$. The answer is (E).