What value of p will make (x$^2$ - 4x + p) a perfect square?

Analyze the question: We want x² - 4x + p to be a perfect square, meaning it can be written as (x - a)² for some number a.

Concept: Completing the square. (x - a)² = x² - 2ax + a²

Solution: Compare x² - 4x + p with x² - 2ax + a²:

• Coefficient of x: -2a = -4, so a = 2
• Constant term: p = a² = 2² = 4

Shortcut: p = (half of the coefficient of x)² = (-4/2)² = (-2)² = 4

Check: x² - 4x + 4 = (x - 2)² ✓

The other options fail:

• p = -2: x² - 4x - 2 has discriminant 16 + 8 = 24, which is not 0
• p = 16: discriminant 16 - 64 = -48, not 0
• p = -8: discriminant 16 + 32 = 48, not 0

A quadratic is a perfect square only when its discriminant b² - 4ac = 0. Here 16 - 4p = 0, so p = 4.

Tip: For x² + bx + p to be a perfect square, always use p = (b/2)². Halve the middle coefficient, then square it. The sign of b does not matter, because squaring makes the result positive.