A problem by Fazla Rabbi

Let $f(x) = x^{19} + x^{11}.$ Note that $f'(x) = 19x^{18} + 11x^{10},$ which is always positive when $x \neq 0.$ Then, besides the single outlier $x = 0,$ $f(x)$ is always increasing. Thus, there can be at most $1$ solution to the given equation.

Since $f(1) = 2 < 9$ and $f(2) = 2^{19} + 2^{11} > 9,$ there must be a root between $x = 1$ and $x = 2.$ Thus there is exactly $\boxed{1}$ root.

Obviously, this polynomial doesn't factor, so we have to find another way to find the number of roots.

If you set $f(x)=x^{19}+x^{11}-9$ , you can find the derivative of $f(x)$ to be $f'(x)=19x^{18}+11x^{10}$ . $f'(x)$ is always positive or $0$ , so $f(x)$ is always increasing. Because of this, $f(x)$ will cross every line $y=k$ exactly once. Setting $k=0$ means that $f(x)$ will cross the $x$ -axis $\boxed{1}$ time.

By Intermediate Value Theorem, we know that there exist a solution in the interval $(1,2)$

Now suppose there's at least two real solutions, call it $a, b$ for $a<b$

Let $f(x) = x^{19} + x^{11} - 9 = 0$ , then for $x < 0$ , $f(x) < 0$ , which means $0 < a < b$

By Fermat's Theorem, there exist a value $c$ in $(a,b)$ such that $f'(c) = 0$ . Then $19c^{18} + 11c^{10} = 0 \Rightarrow c = 0$

Because $a < c < b$ and $c = 0$ , then $a < 0$ which contradicts the above statement.

Hence, there is only $\boxed{1}$ real solution.