Algebra under the condition a + b + c = 0: Find the multiplier k such that a^2 + b^2 + c^2 = k * (ab + bc + ca).

Question

Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:This question checks a well-known identity that holds when the linear sum a + b + c equals zero. It ties the sum of squares to the sum of pairwise products, a relationship that often appears in symmetric algebra and factorization problems. Given Data / Assumptions: a + b + c = 0 We seek k such that a^2 + b^2 + c^2 = k * (ab + bc + ca). Concept / Approach:Start with (a + b + c)^2. Expanding yields a^2 + b^2 + c^2 + 2(ab + bc + ca). Under the given condition, the left side is 0, which directly relates the two target sums and allows k to be read off immediately after rearrangement. Step-by-Step Solution: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).Given a + b + c = 0 ⇒ 0 = a^2 + b^2 + c^2 + 2(ab + bc + ca).Thus a^2 + b^2 + c^2 = −2(ab + bc + ca).Therefore k = −2. Verification / Alternative check: Pick a simple triple with sum zero, e.g., a = 1, b = 1, c = −2. Then a^2 + b^2 + c^2 = 1 + 1 + 4 = 6 and ab + bc + ca = 1 − 2 − 2 = −3. Indeed 6 = (−2) * (−3). Why Other Options Are Wrong: 2, 1, 0: None satisfy the derived identity for generic nonzero triples with a + b + c = 0. Common Pitfalls: Losing the sign on the 2(ab + bc + ca) term during rearrangement. Assuming k must be positive; here it is negative. Final Answer: -2

Algebra under the condition a + b + c = 0: Find the multiplier k such that a^2 + b^2 + c^2 = k * (ab + bc + ca).

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