Equations & Inequations

$\begin{aligned} \sqrt{3x^2+6x+7} + \sqrt{5x^2+10x+14} & \le 4-2x - x^2 \\ \sqrt{3 {\color{#3D99F6}(x+1)^2} +4} + \sqrt{5 {\color{#3D99F6}(x+1)^2} +9} & \le 5- {\color{#3D99F6}(x+1)^2} \end{aligned}$

Note that ${\color{#3D99F6}(x+1)^2} \ge 0$ . Therefore, $\sqrt{3x^2+6x+7}$ and $\sqrt{5x^2+10x+14}$ , and hence the LHS are minumum when $x=-1$ . And the mininum of the LHS $= \sqrt 4 + \sqrt 9 = 5$ . Similarly, the maximum of the RHS $5-(x+1)^2$ also occurs when $x=-1$ and its value is 5. We note that the LHS $>$ RHS when $x \ne -1$ and LHS $=$ RHS when $x=-1$ . Therefore, there is only $\boxed{1}$ solution.