The number of distinct real roots of the equation $$|x||x + 2| - 5|x + 1| - 1 = 0$$ is ______

Let $$y = x+1$$, then $$x = y-1$$. The equation becomes:

$$|(y-1)(y+1)| - 5|y| - 1 = 0 \implies |y^2 - 1| - 5|y| - 1 = 0$$.

Since this is symmetric ($$y$$ and $$-y$$ give same result), let $$t = |y| \ge 0$$:

$$|t^2 - 1| - 5t - 1 = 0$$.

• Case 1: $$t \ge 1$$

$$t^2 - 1 - 5t - 1 = 0 \implies t^2 - 5t - 2 = 0$$.

$$t = \frac{5 \pm \sqrt{25 + 8}}{2} = \frac{5 + \sqrt{33}}{2}$$ (only positive root).

This gives 2 values for $$y$$ (and thus 2 for $$x$$).

• Case 2: $$0 \le t < 1$$

$$-(t^2 - 1) - 5t - 1 = 0 \implies -t^2 + 1 - 5t - 1 = 0 \implies t^2 + 5t = 0$$.

$$t = 0, -5$$. Since $$t = |y|$$, we take $$t = 0$$.

This gives 1 value for $$y$$ ($$y=0$$, so $$x=-1$$).

Total distinct roots: $$2 + 1 = 3$$.

Correct Answer: 3