The number of real roots of the equation $$\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{4}$$ is:

For the equation $$\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{4}$$, we first determine the domain.

For $$\tan^{-1}\sqrt{x(x+1)}$$ to be defined (and real-valued with the argument non-negative), we need $$x(x+1) \ge 0$$, so $$x \le -1$$ or $$x \ge 0$$.

For $$\sin^{-1}\sqrt{x^2+x+1}$$ to be defined, we need $$0 \le x^2+x+1 \le 1$$, i.e., $$x^2+x \le 0$$, which means $$-1 \le x \le 0$$.

Combining both conditions, the domain is $$\{-1, 0\}$$.

Now note that $$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \ge \frac{3}{4}$$ for all real $$x$$. On the domain $$\{-1, 0\}$$: at $$x = 0$$, $$x^2+x+1 = 1$$; at $$x = -1$$, $$x^2+x+1 = 1$$.

At $$x = 0$$: $$\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \ne \frac{\pi}{4}$$.

At $$x = -1$$: $$\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \ne \frac{\pi}{4}$$.

Neither point in the domain satisfies the equation. Furthermore, since $$\sin^{-1}\sqrt{x^2+x+1} \ge \sin^{-1}(1) = \frac{\pi}{2}$$ at the boundary points (where $$x^2+x=0$$, giving the value 1 under the square root), the left side is always at least $$\frac{\pi}{2} \gt \frac{\pi}{4}$$ throughout the domain.

Therefore the equation has $$0$$ real roots.