Both $|x-4|$ and $\sqrt{9-x}$ are $\ge 0$. Once you restrict $x$ so that $9 - x \ge 0$, you can square both sides to remove the absolute value and the square root. What equation do you get after squaring and simplifying?

After squaring and simplifying, you get a quadratic equation. You can solve it with the quadratic formula, but to find the product of its solutions more quickly, recall that for $ax^2 + bx + c = 0$, the product of the roots is $\dfrac{c}{a}$.

Solving the quadratic gives candidate solutions. Make sure each one actually satisfies the original equation with the absolute value and square root before including it in the product.

In Desmos, enter y = abs(x - 4) on one line and y = sqrt(9 - x) on another line. These represent the left and right sides of the equation.

Zoom or pan the graph so both curves are clearly visible. Tap/click where the graphs intersect, or use Desmos’s intersection tool, to find the coordinates of all intersection points; note the x-values of these points, since they are the real solutions.

In a new Desmos expression line, type the product of the intersection x-values (for example, if they are $x=a$ and $x=b$, type a*b). Desmos will display the numerical value of this product; compare that value with the answer choices.

Because of the square root, the expression inside it must be nonnegative:

• $9 - x \ge 0$, so $x \le 9$.

Also, both $|x-4|$ and $\sqrt{9-x}$ are always $\ge 0$, so we can square both sides (we will still check solutions afterward):

which simplifies to

Now expand $(x-4)^2$ and move all terms to one side:

Add $x$ to both sides and subtract $9$ from both sides:

So any solution of the original equation must satisfy the quadratic

Use the quadratic formula on $x^2 - 7x + 7 = 0$:

So the quadratic has two real roots:

• $x_1 = \dfrac{7 + \sqrt{21}}{2}$
• $x_2 = \dfrac{7 - \sqrt{21}}{2}$

These are the candidate solutions for the original equation.

We must make sure these values actually satisfy $|x-4| = \sqrt{9-x}$.

First note:

• $\sqrt{21} \approx 4.58$, so $x_1 \approx \dfrac{7 + 4.58}{2} \approx 5.79$ (which is $> 4$ and $< 9$).
• $x_2 \approx \dfrac{7 - 4.58}{2} \approx 1.21$ (which is $< 4$ and $< 9$).

Now check each approximately in the original equation:

• For $x_1 \approx 5.79$:
  • $|x_1 - 4| \approx |1.79| \approx 1.79$
  • $\sqrt{9 - x_1} \approx \sqrt{3.21} \approx 1.79$
• For $x_2 \approx 1.21$:
  • $|x_2 - 4| \approx |-2.79| \approx 2.79$
  • $\sqrt{9 - x_2} \approx \sqrt{7.79} \approx 2.79$

In both cases, the left and right sides match, so both $x_1$ and $x_2$ are real solutions of the original equation.

The equation $x^2 - 7x + 7 = 0$ has solutions $x_1$ and $x_2$, and we have confirmed both are valid.

For a quadratic $ax^2 + bx + c = 0$, the product of the roots is $\dfrac{c}{a}$. Here $a = 1$ and $c = 7$, so

Therefore, the product of all real solutions to $|x-4| = \sqrt{9-x}$ is $\boxed{7}$, which corresponds to choice C.