The polynomial $f(x)$ is given by $f(x)=9x^3-30x^2-23x-4$.

1. Use the Factor Theorem to find a linear factor of $f(x)$.

Let $f(x) = 9x^3 - 30x^2 - 23x - 4$.
Testing integer factors of $-4$:
$f(4) = 9(4)^3 - 30(4)^2 - 23(4) - 4$
$f(4) = 576 - 480 - 92 - 4 = 0$
By the Factor Theorem, $(x-4)$ is a factor of $f(x)$.

2. Find the quadratic factor by comparing coefficients or long division.

Let $9x^3 - 30x^2 - 23x - 4 = (x-4)(ax^2 + bx + c)$.
By comparing the coefficients of $x^3$ and the constant term:
$a = 9$
$-4c = -4 \implies c = 1$
So, $9x^3 - 30x^2 - 23x - 4 = (x-4)(9x^2 + bx + 1)$.
Comparing the coefficients of $x^2$:
$-36 + b = -30 \implies b = 6$
Thus, the quadratic factor is $9x^2 + 6x + 1$.

3. Factorise the quadratic factor completely.

$9x^2 + 6x + 1 = (3x+1)^2$
Therefore, $f(x) = (x-4)(3x+1)^2$.

1. Rearrange the given equation and substitute $u = \sqrt{x}$ to form a polynomial equation.

$9x\sqrt{x} - 30x - 23\sqrt{x} - 4 = 0$
Let $u = \sqrt{x}$. Then $x = u^2$ and $x\sqrt{x} = u^3$.
$9u^3 - 30u^2 - 23u - 4 = 0$
This is exactly $f(u) = 0$.

2. Use the factorised form from part (a) to solve for $u$.

From part (a), we know $f(x) = (x-4)(3x+1)^2$.
Therefore, $f(u) = (u-4)(3u+1)^2 = 0$.
$u = 4 \quad \text{or} \quad u = -\frac{1}{3}$

3. Solve for $x$ and reject the invalid root to prove there is only one real root.

Since $u = \sqrt{x}$, we have:
$\sqrt{x} = 4 \quad \text{or} \quad \sqrt{x} = -\frac{1}{3}$
The principal square root $\sqrt{x}$ must be non-negative, so $\sqrt{x} = -\frac{1}{3}$ has no real solution.
Thus, the only real root comes from $\sqrt{x} = 4$.
$x = 4^2 = 16$
This proves the equation has only one real root, which is $x = 16$.