If r a constant such that $|x^2 - 4x - 13| = r$ has exactly three distinct real roots, then the value of r is

We need to find the value of r such that $|x^2 - 4x - 13| = r$ has exactly three distinct real roots.

The equation $|f(x)| = r$ represents the intersection of a horizontal line $y = r$ with the graph of $|f(x)|$. For exactly 3 intersections, something special must happen!

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Let's work with $f(x) = x^2 - 4x - 13$

Converting to vertex form by completing the square:

$f(x) = x^2 - 4x - 13$

$f(x) = (x^2 - 4x + 4) - 4 - 13$

$f(x) = (x - 2)^2 - 17$

The vertex form tells us the parabola has its minimum point at $(2, -17)$.

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Since $f(x) = (x - 2)^2 - 17$:

The parabola opens upward (coefficient of $x^2$ is positive)

Minimum value occurs at $x = 2$, where $f(2) = -17$

The parabola dips below the x-axis (since minimum is $-17 < 0$)

For $|f(x)|$:

When $f(x) \geq 0$: $|f(x)| = f(x)$ (the graph stays the same)

When $f(x) < 0$: $|f(x)| = -f(x)$ (the graph flips above the x-axis)

This creates a "W" shaped graph where the bottom part of the parabola is reflected upward.

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The equation $|f(x)| = r$ means we're looking for where a horizontal line $y = r$ intersects the "W" shaped graph.

The lowest point of the "W" graph occurs at $x = 2$, where $|f(2)| = |-17| = 17$.

Why $r = 17$ gives exactly 3 roots:

When $r = 17$: The horizontal line $y = 17$ touches the bottom tip of the "W" at exactly one point $(x = 2)$

This same line intersects the two upper branches of the "W" at exactly two more points

Total: 3 intersection points = 3 roots

For other values of r:

If $r < 17$: No intersections (0 roots)

If $r > 17$: Four intersections (4 roots)

If $r = 17$: Exactly 3 intersections

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$r = 17$

For $|f(x)| = r$ to have exactly 3 roots, r must equal the absolute value of the minimum value of $f(x)$. This creates the special case where the horizontal line just touches the bottom of the "W" shaped graph while intersecting the upper branches.