Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$ Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$If $k,w,f$ are relatively prime positive integers,then find the value of $k+w+f.$

The given equation is $\log_8(kx^2+wx+f)=2$ i.e. $kx^2+wx+f=64$
Since $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,so
$\frac{k}{9}+\frac{w}{3}+f=64.............(1)$
$225k-15w+f=64............................(2)$
I need to find $k,w,f$ from these equations,but i need three equations to find three variables.I am stuck here.

Eliminating $f$ gives $44k=3w$. Take $w=44,k=3$. Then we get $f=49$. Note that the general solution is $k=3h,w=44h,f=64-15h$, but the requirement that the numbers are relatively prime positive integers forces $h=1$.
