2025 · Q27
If (a-b)^2 + (b-c)^2 + (c-a)^2 = 6 and a^2+b^2+c^2 = 29, then what is the value of a+b+c?
Elementary MathematicsAlgebra
- A9
- B-9
- C±9
- D0
Explanation
Expanding, (a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ca)=6 gives ab+bc+ca=26. Then (a+b+c)^2=29+52=81, so a+b+c=+/-9 mathematically - both signs are realizable with real a,b,c. The stored key and the published coaching key both say A (9), suggesting the original question constrained a,b,c (e.g. positive), so the stored key is kept. Note: key disputed - independent verification inconclusive.