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.

At $x_0=\frac{1/3+(-15)}{2}$ there will be an extermum, since the value of the function at these two points are the same. Hence the derivative of the quadratic function, which is $2kx+w$ should be zero at $x_0$. That gives you the third equation.
