2. If the solution set of the inequality $\sqrt{x}>a x+\frac{3}{2}$ is $(4, b)$, then the real number $a=$ $\qquad$ , $b=$ $\qquad$ -
Solve for $\frac{1}{8}, 36$.
Method 1 Let $\sqrt{x}=t$, then $x=t^{2}$, and $t \in(2, \sqrt{b})$, so the solution set of the inequality $a t^{2}-t+\frac{3}{2}a x+\frac{3}{2}$ is $(4, b)$. This means the graphs of the two functions $y_{1}=\sqrt{x}, y_{2}=$ $a x+\frac{3}{2}$ intersect in the same coordinate system.
Let the intersection points of the two function graphs be $A, B$, then $A(4,2), B(b, \sqrt{b})$, so $2=4 a+\frac{3}{2}, \sqrt{b}=a b+\frac{3}{2}$, solving these gives $a=\frac{1}{8}, b=36$.