Find all real values of p for which the difference between the roots of x^2 + p x + 8 = 0 equals 2.

Introduction / Context: The spacing of roots of a monic quadratic can be related directly to its coefficients using Vieta’s relations and the identity |α − β| = sqrt((α + β)^2 − 4αβ). Here, the spacing is fixed at 2; we solve for the parameter p that enforces this.

Given Data / Assumptions:

• Equation: x^2 + p x + 8 = 0.
• |α − β| = 2.
• We seek real p values.

Concept / Approach: For monic quadratics, α + β = −p and αβ = 8. Then |α − β| = sqrt((α + β)^2 − 4αβ) = sqrt(p^2 − 32). Set this equal to 2 and solve.

Step-by-Step Solution:

Verification / Alternative check: With p = 6 or p = −6, compute the discriminant and verify the roots’ difference equals 2; both values work.

Why Other Options Are Wrong: ±2, ±4, ±8 do not satisfy p^2 − 32 = 4; “No real p” contradicts the solution.

Common Pitfalls: Forgetting the absolute value in the difference or squaring both sides incorrectly.

Final Answer: p = ±6