A geometry problem by Aly Ahmed

Let $x^2+x=t$ then $\displaystyle \tan^{-1}\sqrt{t}=\frac{\pi}{2}-\sin^{-1}\sqrt{t+1}=\cos^{-1}\sqrt{t+1}$

Now , $\displaystyle \cos^{-1}\sqrt{t+1}=\tan^{-1}\sqrt{\frac{-t}{t+1}}$ and since $\displaystyle \frac{-t}{t+1}>0$ so $-1<t<0$

Therefore , $\displaystyle \tan^{-1}\sqrt{t}=\tan^{-1}\sqrt{\frac{-t}{t+1}}\implies t=\frac{-t}{t+1} \implies t+2=0$

So $\displaystyle x^2+x+2=0$ and sum of roots is $\boxed{-1}$

$\begin{aligned} \tan^{-1} \sqrt{x(x+1)} + \sin^{-1} \sqrt{x^2+x+1} & = \frac \pi 2 \\ \tan^{-1} \sqrt{x(x+1)} & = \frac \pi 2 - \sin^{-1} \sqrt{x^2+x+1} \\ \sin \left(\tan^{-1} \sqrt{x(x+1)} \right) & = \sin \left(\frac \pi 2 - \sin^{-1} \sqrt{x^2+x+1} \right) \\ \frac {\sqrt{x(x+1)}}{\sqrt{x^2+x+1}} & = \sqrt{-x^2-x} \\ \frac {x^2+x}{x^2+x+1} & = -x^2-x \\ x^2+x+1 & = -1 \\ x(x+1) & = 0 \\ \implies x & = \begin{cases} 0 \\ -1 \end{cases} \end{aligned}$

Therefore, the sum of roots of the equation is $0-1= \boxed{-1}$