Andrew's function - 2

Putting $x=1,2$ in $A(x)=px+q^x$ , we have $A(1)=p+q=4$ and $4A(2)=8p+4q^2=37$ and substituting $p=4-q$ in the second equation we have $8(4-q)+4q^2=37 \Rightarrow 4q^2-8q-5=0$ . Solving this quadratic we have $q=\dfrac{5}{2} \ , \ -\dfrac{1}{2}$ and the corresponding values of $p= \dfrac{3}{2} \ , \ \dfrac{9}{2}$ . So for $p-q$ we have two values $5$ and $-1$ and the maximum value out of these is $\boxed{5}$ .