Let $f(x)=\sqrt{\frac{x^2+ax+4}{x^2+bx+16}}$ is defined for all real $x$,then find the number of possible ordered pairs $(a,b),$

Question

Let $f(x)=\sqrt{\frac{x^2+ax+4}{x^2+bx+16}}$ is defined for all real $x$,then find the number of possible ordered pairs $(a,b),$ where $a,b$ are both integers.

---

As $f(x)$ is defined for all real $x$,

so $\frac{x^2+ax+4}{x^2+bx+16}\geq0$ and $x^2+bx+16\neq0$
I do not know how to proceed further.

Answer 1

The function is defied for all real $x$, so the denominator must not equal $0$, its discriminant is negative, i.e. $b^2-64<0$

Notice the deniminator must be positive, for the square root to exist, the numerator must be positive or $0$ too. So discriminant of it is smaller than or equal to zero. $a^2 -16\leq 0$

Answer 2

$\left|b\right| \lt 8$. For any other choice of $b$, the denominator has real roots and then the RHS is undefined for some real $x$. Also $x^2+bx+16\gt 0$ for $b$ in the specified range.

For $f(x)$ to be well defined, $x^2+ax+4\ge 0$. This is again guaranteed if $\left|a\right|\le 4$.