Find all real values of $a$ for which the equation $(x^2 + a)^2 + a = x$ has four real roots.
I played around a bit with a graphing calculator and suspect that $(-\infty, -\frac{3}{4})$ are solutions, but I'm not sure if that's all.
2026-04-11 10:48:08.1775904488
Tough polynomial form problem
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We have the cheeky difference of squares factorization : $$ (x^2+a)^2 + a-x = (\color{red}{x^2+a})^2 -\color{blue}{x}^2 +x^2+a-x \\ = (\color{red}{x^2+a}-\color{blue}{x})(\color{red}{x^2+a}+\color{blue}x)+x^2+a-x = (x^2+x+a+1)(x^2-x+a) $$
which has four real roots if and only if both the factors have two real roots each, which is if and only if their discriminants are non-negative. Find these, solve the simple inequalities for $a$ and combine the ranges.