1. The equation $\sqrt{a x^{2}+a x+2}=a x+2$ has a unique real root, then the range of the real number $a$ is $\qquad$
1. $a=-8$ or $a \geq 1$

Analysis: The original equation can be transformed into $\left(a^{2}-a\right) x^{2}+3 a x+2=0$ and $a x+2 \geq 0$. (1) When $a^{2}-a=0$, i.e., $a=0$ or 1, it is easy to see that $a=1$ meets the requirement;
(2) When $a^{2}-a \neq 0$, i.e., $a \neq 0,1$, (i) $\triangle=a^{2}+8 a=0$, i.e., $a=-8$, meets the requirement; (ii) $\triangle=a^{2}+8 a>0$, i.e., $a0$ when it is easy to see that $-\frac{2}{a}$ is not a solution of the equation, thus $x_{1}1$
Therefore, the range of $a$ is $a=-8$ or $a \geq 1$.