If \(a^2+b^2+c^2=6.25\) and \(ab+bc+ca=0.52\), find \(a+b+c\), given that \(a+b+c<0\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-2.7\)
The square of a sum of three terms is \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\). The square root gives two signs, and the condition \(a+b+c<0\) keeps only the negative one.
\((a+b+c)^2=6.25+2\times0.52=6.25+1.04=7.29\)
\(\sqrt{7.29}=2.7\), so \(a+b+c=2.7\) or \(-2.7\)
\(a+b+c<0\), so \(a+b+c=-2.7\)
Why the other options are incorrect
Option (1) \(\pm2.7\) — Incorrect. The value \(+2.7\) is ruled out by the condition \(a+b+c<0\).
Option (3) \(\pm2.8\) — Incorrect. \(2.8^2=7.84\), which is not 7.29; it also keeps a positive value.
Option (4) \(-2.8\) — Incorrect. \((-2.8)^2=7.84\), not 7.29.