Opposite-sign, equal-magnitude roots condition: For px^2 + qx + r = 0 with real coefficients and p ≠ 0, when are the two roots equal in magnitude but opposite in sign?

Introduction / Context:Having roots equal in magnitude and opposite in sign means the roots are a and −a (a ≠ 0). For a quadratic px^2 + qx + r = 0, this imposes simple conditions on the coefficients via Vieta’s formulas. We translate the root pattern into constraints on sum and product and match with the options.

Given Data / Assumptions:

• Roots are a and −a with a ≠ 0 (so they are nonzero and opposite in sign).
• Sum of roots = −q/p
• Product of roots = r/p

Concept / Approach:For roots a and −a, the sum is zero, hence −q/p = 0 ⇒ q = 0. The product is (a)(−a) = −a^2 < 0, so r/p < 0. That implies p and r are nonzero and of opposite signs, i.e., pr < 0 (in particular pr ≠ 0). Among the provided choices, the one that enforces q = 0 and keeps pr nonzero (hence allowing opposite signs) is the best match.

Step-by-Step Solution:

Verification / Alternative check:Example: Take p = 1, r = −4, q = 0. Equation x^2 − 4 = 0 has roots ±2: equal magnitude, opposite signs. This matches the condition.

Why Other Options Are Wrong:

• q = 0, r = 0 gives a repeated zero root, not opposite-signed nonzero roots.
• p = 0 is not a quadratic.
• r = 0 implies one root is zero (not ±a with a ≠ 0).

Common Pitfalls:Choosing q = 0, r = 0 (tempting but wrong). Opposite-signed, equal-magnitude roots require nonzero product of negative sign.

Final Answer:q = 0, pr ≠ 0