Algebra under the condition a + b + c = 0: Find the multiplier k such that a^2 + b^2 + c^2 = k * (ab + bc + ca).
Aptitude
Elementary Algebra
Difficulty: Easy
Choose an option
-
A-2
-
B0
-
C1
-
D2
Answer
Correct Answer: -2
Explanation
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