${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$ For equation to be defined, x 2 + x $\ge$ 0 $\Rightarrow$ x 2 + x + 1 $\ge$ 1 $\therefore$ Only possibility that the equation is defined x 2 + x = 0 $\Rightarrow$ x = 0; x = $-$1 None of these values satisfy $\therefore$ No of roots = 0