7. Let $a, b>0$, satisfy: the equation $\sqrt{|x|}+\sqrt{|x+a|}=b$ has exactly three distinct real solutions $x_{1}, x_{2}, x_{3}$, and $x_{1}<x_{2}<x_{3}=b$, then the value of $a+b$ is $\qquad$ .
Let $t=x+\frac{a}{2}$, then $\sqrt{|x|}+\sqrt{|x+a|}=\sqrt{\left|t-\frac{a}{2}\right|}+\sqrt{\left|t+\frac{a}{2}\right|}$. Note that the function $f(t)=\sqrt{\left|t-\frac{a}{2}\right|}+\sqrt{\left|t+\frac{a}{2}\right|}$ is an even function and $f(t)=b$ has exactly three distinct real solutions, so $f(0)=b$. Solving the system of equations $\left\{\begin{array}{l}\sqrt{|b|}+\sqrt{|a+b|}=b, \\ \sqrt{\left|\frac{a}{2}\right|}=\frac{b}{2}\end{array} \Rightarrow\left\{\begin{array}{l}\sqrt{b}+\sqrt{a+b}=b, \\ b=\sqrt{2 a}\end{array}\right.\right.$ $\Rightarrow \sqrt{b}+\sqrt{\frac{b^{2}}{2}+b}=b \Rightarrow 1+\sqrt{\frac{b}{2}+1}=\sqrt{b} \Rightarrow \frac{b}{2}+1=b-2 \sqrt{b}+1 \Rightarrow b=16$, so $a=\frac{b^{2}}{2}=128 \Rightarrow a+b=144$.