Find all real values of p for which the difference between the roots of x^2 + p x + 8 = 0 equals 2.
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A± 2
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B± 4
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C± 6
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D± 8
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ENo real p
Answer
Correct Answer: ± 6
Explanation
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:
sqrt(p^2 − 32) = 2 ⇒ p^2 − 32 = 2^2 = 4.p^2 = 36 ⇒ p = ±6.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