Solve it without calculus or quadratic

Yash has provided an elegant solution, so I'll provide the 'boring' one.

Let $|AD| = d,$ and let $v$ be the person's speed on the highway. Then the time of travel from $A$ to $C$ is given by the equation

$t = \dfrac{d - x}{v} + \dfrac{\sqrt{x^{2} + l^{2}}}{\dfrac{v}{\sqrt{10}}}.$

Now since $d$ and $l$ are constants, the derivative with respect to $x$ is

$\dfrac{dt}{dx} = -\dfrac{1}{v} + \dfrac{\sqrt{10}*x}{v\sqrt{x^{2} + l^{2}}}.$

To find any extrema, we set this equal to $0,$ which gives us

$\sqrt{x^{2} + l^{2}} = \sqrt{10}*x \Longrightarrow x^{2} + l^{2} = 10x^{2} \Longrightarrow 9x^{2} = l^{2} \Longrightarrow x = \dfrac{l}{3}.$

So with $l = 9$ m we end up with $x = \boxed{3}$ m.

To demonstrate that this value yields a minimum, note that

$\dfrac{d^{2}t}{dx^{2}} = \dfrac{\sqrt{10}*l^{2}}{v(x^{2} + l^{2})^{\frac{3}{2}}} \gt 0,$

and thus by the second derivative test any extrema will be a minimum.

Refractive index method is not applied as the case is not usual.

Like Yash I solved this using the analogy with optics.

I imagined the light ray going from C, through B to A.

The refracted ray BA lies along the boundary.

So CBD is the critical angle which is related to the refractive index $\sqrt{10}$ by

$\sin{CBD}=\dfrac{1}{\sqrt{10}}$

From this it follows easily that

$\tan{CBD}=\dfrac{1}{3}$

and so

$BD=\dfrac{9}{3}=\boxed{3}$