Value of a fraction independent of x

Question

For $a=4$ it is known that the value of the fraction $\frac{(a+2)x + a^2 - 1}{ax-2a + 18}$ is independent of $x$. The other value of $a$ for which this is the case, belong to the interval _______.

My approach : Since the fraction is independent of x for $a=4$, so we can assume that the value becomes $\frac{15}{10}$ = $\frac{3}{2}$, by substituting $x = 0$. I don't know how to approach to the next step after this. Please help.

Answer 1

If $a=4$, then your fractions is just $\frac32$. Is there another $a$ for which the fractions is constant? That would mean that $\frac{a+2}a=\frac{a^2-1}{-2a+18}$, which is equivalent to $a^3+2a^2-15-36=0$. This is a cubic equation, but you already know that it has $4$ as a root. The only other root is $-3$.

Answer 2

Hint

Consider the function $$f(x)=\frac{(a+2)x + a^2 - 1}{a(x-2) + 18}$$ Compute its derivative to get $$f'(x)=-\frac{(a-4) (a+3)^2}{(a (x-2)+18)^2}$$

IU am sure that you can take it from here.