For each quadratic that you get, think about using the discriminant $b^2 - 4ac$ to decide whether it has 0, 1, or 2 real solutions, instead of solving for the exact roots.

The total number of solutions to the absolute value equation is the number of distinct roots from both quadratics. Ask yourself: for which values of $k$ could a number $x$ possibly satisfy both quadratic equations at the same time?

For each value of $k$ in the choices, determine how many real roots each quadratic has and whether any roots coincide. Then look for the choice where the total number of distinct real solutions is exactly three.

In Desmos, type y = abs(x^2 - 4x) to graph the function $y = |x^2 - 4x|$. Adjust the viewing window so you can clearly see where the graph crosses and curves near the x-axis.

For each answer choice, add a horizontal line: type y = 0, y = 2, y = 4, and y = 8 in separate lines (you can toggle them on and off). Each line represents $|x^2 - 4x| = k$ for that particular $k$.

For each horizontal line, look at how many points it intersects the curve $y = |x^2 - 4x|$. Use the intersection tool (tap the points where the graphs meet) to see how many distinct x-values you get for each $k$, and identify which $k$ gives exactly three intersection points.

For any expression $A$ and nonnegative number $k$, the equation $|A| = k$ is equivalent to $A = k$ or $A = -k$.

Here $A = x^2 - 4x$, so

That gives two quadratic equations to consider:

The total number of solutions to $|x^2 - 4x| = k$ will be the number of distinct real roots these two equations have together.

For a quadratic $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$:

• If $D > 0$, there are 2 distinct real roots.
• If $D = 0$, there is 1 real root (a repeated root).
• If $D < 0$, there are no real roots.

Apply this to each equation.

1. For $x^2 - 4x - k = 0$:
   • $a = 1$, $b = -4$, $c = -k$.
   • Discriminant:

• For all answer choices $k \ge 0$, $16 + 4k > 0$, so this equation always has 2 distinct real roots.

2. For $x^2 - 4x + k = 0$:
   • $a = 1$, $b = -4$, $c = k$.
   • Discriminant:

• Here, the number of real roots depends on $k$:
  • If $16 - 4k > 0$ (i.e., $k < 4$), then 2 distinct real roots.
  • If $16 - 4k = 0$ (i.e., $k = 4$), then 1 real root (a repeated root).
  • If $16 - 4k < 0$ (i.e., $k > 4$), then no real roots.

We must be careful not to double-count any roots that might solve both quadratics.

If some $x$ satisfies both

then subtracting the equations gives

So the two quadratics can share roots only when $k = 0$.

Now use this and the discriminant information from Step 2 for each choice:

• $k = 0$:

  • Both equations become $x^2 - 4x = 0$, so they are the same. Solving gives $x(x - 4) = 0$, so $x = 0$ or $x = 4$ → 2 distinct solutions total.

• $k = 2$ (note $k < 4$):

  • First equation: always 2 distinct roots.
  • Second equation: $16 - 4k = 16 - 8 = 8 > 0$ → 2 distinct roots.
  • Since $k \ne 0$, these roots are all different → 4 distinct solutions total.

• $k = 4$:

  • First equation: still 2 distinct roots (because $16 + 4k = 32 > 0$).
  • Second equation: $16 - 4k = 16 - 16 = 0$ → exactly 1 real root (a repeated root).
  • Because shared roots can only happen when $k = 0$, this 1 root is different from the 2 roots of the first equation.
  • So we get 2 + 1 = 3 distinct real solutions.

• $k = 8$ ($k > 4$):

  • First equation: 2 distinct roots.
  • Second equation: $16 - 4k = 16 - 32 = -16 < 0$ → no real roots.
  • Total: 2 distinct solutions.

Only $k = 4$ makes the original equation $|x^2 - 4x| = k$ have exactly three distinct real solutions, so the correct answer is 4.